Welcome to HSLC Guru! This page presents the complete ASSEB Class 10 Mathematics Chapter 7 Question Answer | Coordinate Geometry in English Medium. Coordinate Geometry is a beautiful bridge between algebra and geometry: every point in a plane is described by an ordered pair of numbers, and every geometric relationship — distance, division, area, collinearity — becomes an equation. This chapter unlocks three powerful tools: the Distance Formula, the Section Formula (including the midpoint), and the Area of a Triangle. We work through every problem of Exercises 7.1, 7.2, 7.3 and 7.4 of the ASSEB textbook with clear coordinate-plane diagrams, step-by-step algebra and ready-to-revise summaries. Whether you want quick revision before the HSLC examination or deep mastery of every concept, scroll on — each formula is paired with worked examples and additional practice questions to make the chapter unforgettable.

Chapter Summary

The plane is divided into four quadrants by two perpendicular axes — the X-axis (horizontal) and the Y-axis (vertical). Their intersection is the origin $O(0,0)$. Any point $P$ in the plane is represented by an ordered pair $(x, y)$ where $x$ is the abscissa (perpendicular distance from Y-axis) and $y$ is the ordinate (perpendicular distance from X-axis). Signs of coordinates in each quadrant are: I $(+,+)$, II $(-,+)$, III $(-,-)$, IV $(+,-)$. Points on the X-axis have ordinate $0$ — written $(x, 0)$; points on the Y-axis have abscissa $0$ — written $(0, y)$.

Coordinate Geometry was developed by the French mathematician and philosopher René Descartes (1596-1650) in the seventeenth century — which is why the rectangular coordinate system also bears the name Cartesian system. The fundamental insight is that points in the plane correspond one-to-one with ordered pairs of real numbers. This allows every geometric statement (lengths, ratios, areas, parallelism, perpendicularity) to be translated into algebraic equations that can be manipulated and solved. Conversely, every algebraic equation in two variables defines a curve, and the algebraic properties of the equation tell us geometric facts about the curve. This back-and-forth between algebra and geometry is one of the most powerful ideas in all of mathematics, and it underlies modern subjects from physics and computer graphics to data science and engineering.

In Class 9 you learnt how to plot a point given its coordinates, and how to read off the coordinates of a plotted point. In this chapter we go further — given the coordinates of two or more points, we compute meaningful quantities about the figure they form: the distance between them, a special point that splits the segment in a chosen ratio, the area of a triangle, and a test for whether three points lie on a single straight line.

Building on this skeleton, the chapter answers three big questions:

• How far apart are two points? — solved by the Distance Formula.
• Where does a point divide a segment in a given ratio? — solved by the Section Formula (and the midpoint as a special case).
• What is the area of a triangle whose vertices are known? — solved by the area formula. The same formula identifies collinear points (area $=0$).

Key Formulas to Remember

1. Distance Formula: Distance between $A(x_1,y_1)$ and $B(x_2,y_2)$ is

$$AB = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$$

2. Distance from the Origin: Distance of $P(x,y)$ from $O(0,0)$ is

$$OP = \sqrt{x^2 + y^2}$$

3. Section Formula (internal division): The point $P$ which divides $A(x_1,y_1)$ and $B(x_2,y_2)$ internally in the ratio $m:n$ is

$$P = \left(\frac{m x_2 + n x_1}{m+n},\ \frac{m y_2 + n y_1}{m+n}\right)$$

4. Midpoint Formula: Midpoint of $A(x_1,y_1)$ and $B(x_2,y_2)$ is

$$M = \left(\frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2}\right)$$

5. Area of a Triangle with vertices $A(x_1,y_1)$, $B(x_2,y_2)$, $C(x_3,y_3)$:

$$\text{Area} = \tfrac{1}{2}\left|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)\right|$$

6. Collinearity Test: Three points $A$, $B$, $C$ are collinear if and only if Area of $\triangle ABC = 0$.

7. Centroid of a Triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$:

$$G = \left(\frac{x_1+x_2+x_3}{3},\ \frac{y_1+y_2+y_3}{3}\right)$$

How the Distance Formula is Derived

Suppose we have two points $A(x_1,y_1)$ and $B(x_2,y_2)$. Drop perpendiculars from $A$ and $B$ on the X-axis at $L$ and $M$, and from $A$ a perpendicular on $BM$ meeting it at $N$. Then $AN = LM = x_2 – x_1$ and $BN = y_2 – y_1$. The triangle $ANB$ is right-angled at $N$, so by Pythagoras’ theorem:

$$AB^2 = AN^2 + BN^2 = (x_2-x_1)^2 + (y_2-y_1)^2$$

Taking the positive square root (because distance is always non-negative) gives the distance formula. The squaring step makes the formula symmetric in the two points: it does not matter which point you call $A$ and which $B$ — you get the same answer.

How the Section Formula is Derived

Let $P(x,y)$ divide the segment $AB$ internally in the ratio $m:n$, so that $AP:PB = m:n$. Drop perpendiculars from $A$, $P$, $B$ on the X-axis at $L$, $Q$, $M$ respectively, and from $A$ and $P$ horizontal segments meeting these verticals. By the basic proportionality theorem (the line $PQ$ parallel to $AL$ and $BM$ cuts $AB$ in the same ratio as it cuts $LM$):

$$\frac{LQ}{QM} = \frac{AP}{PB} = \frac{m}{n}$$

Hence $\dfrac{x – x_1}{x_2 – x} = \dfrac{m}{n}$, which on cross-multiplying gives $n(x – x_1) = m(x_2 – x)$, and rearranging:

$$x = \frac{m x_2 + n x_1}{m + n}$$

An identical argument with the Y-coordinates gives $y = \dfrac{m y_2 + n y_1}{m+n}$. Putting $m=n$ (or simply $1:1$) yields the midpoint formula as a special case.

How the Area Formula Works

The area of a triangle with vertices $A(x_1,y_1)$, $B(x_2,y_2)$, $C(x_3,y_3)$ can be obtained by surrounding the triangle with three trapeziums under the X-axis and adding/subtracting their areas. The clean closed form is:

$$\text{Area} = \tfrac{1}{2}|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$$

Two important consequences:

• If the three points are collinear (lie on the same line) the “triangle” collapses, so the area is $0$. We use this as a quick collinearity test.
• The expression inside the modulus may come out negative depending on whether the vertices are listed clockwise or anti-clockwise. The modulus removes that sign because area is always positive.

Exercise 7.1 — Distance Formula

Q1. Find the distance between the following pairs of points:

(i) $(2,3)$ and $(4,1)$

Answer: Using $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:

$$d = \sqrt{(4-2)^2 + (1-3)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \text{ units}$$

(ii) $(-5,7)$ and $(-1,3)$

Answer:

$$d = \sqrt{(-1-(-5))^2 + (3-7)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \text{ units}$$

(iii) $(a,b)$ and $(-a,-b)$

Answer:

$$d = \sqrt{(-a-a)^2 + (-b-b)^2} = \sqrt{4a^2 + 4b^2} = 2\sqrt{a^2+b^2} \text{ units}$$

Q2. Find the distance between the points $(0,0)$ and $(36,15)$. Can you now find the distance between the two towns A and B discussed in Section 7.2?

Answer:

$$d = \sqrt{36^2 + 15^2} = \sqrt{1296 + 225} = \sqrt{1521} = 39 \text{ units}$$

Yes, since the towns lie at $(0,0)$ and $(36,15)$ on the coordinate grid, the distance between them is $39$ km.

Q3. Determine if the points $(1,5)$, $(2,3)$ and $(-2,-11)$ are collinear.

Answer: Let $A(1,5)$, $B(2,3)$, $C(-2,-11)$.

$$AB = \sqrt{(2-1)^2 + (3-5)^2} = \sqrt{1+4} = \sqrt{5}$$

$$BC = \sqrt{(-2-2)^2 + (-11-3)^2} = \sqrt{16+196} = \sqrt{212}$$

$$AC = \sqrt{(-2-1)^2 + (-11-5)^2} = \sqrt{9+256} = \sqrt{265}$$

Since $AB + BC \neq AC$ (and no two sums match), the points are not collinear.

Q4. Check whether $(5,-2)$, $(6,4)$ and $(7,-2)$ are the vertices of an isosceles triangle.

Answer: Let $A(5,-2)$, $B(6,4)$, $C(7,-2)$.

$$AB = \sqrt{(6-5)^2 + (4+2)^2} = \sqrt{1+36} = \sqrt{37}$$

$$BC = \sqrt{(7-6)^2 + (-2-4)^2} = \sqrt{1+36} = \sqrt{37}$$

$$AC = \sqrt{(7-5)^2 + 0} = 2$$

Since $AB = BC = \sqrt{37}$, the triangle is isosceles.

Q5. In a classroom, 4 friends are seated at the points $A(3,4)$, $B(6,7)$, $C(9,4)$ and $D(6,1)$. Champa and Chameli walk into the class and say “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.

Answer:

$$AB = \sqrt{9+9} = 3\sqrt{2},\ \ BC = \sqrt{9+9} = 3\sqrt{2}$$

$$CD = \sqrt{9+9} = 3\sqrt{2},\ \ DA = \sqrt{9+9} = 3\sqrt{2}$$

$$AC = \sqrt{36+0} = 6,\ \ BD = \sqrt{0+36} = 6$$

All four sides are equal and both diagonals are equal, so $ABCD$ is a square. Champa is correct.

Q6. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:

(i) $(-1,-2),(1,0),(-1,2),(-3,0)$

Answer: Let $A(-1,-2)$, $B(1,0)$, $C(-1,2)$, $D(-3,0)$.

$AB = BC = CD = DA = 2\sqrt{2}$. Diagonals: $AC = 4$, $BD = 4$. All sides equal and diagonals equal — $ABCD$ is a square.

(ii) $(-3,5),(3,1),(0,3),(-1,-4)$

Answer: Computing the four distances, three of the points $(-3,5)$, $(0,3)$, $(3,1)$ are collinear (they all satisfy $2x+3y=9$), so no quadrilateral is formed.

(iii) $(4,5),(7,6),(4,3),(1,2)$

Answer: Let $A(4,5)$, $B(7,6)$, $C(4,3)$, $D(1,2)$.

$AB = \sqrt{9+1} = \sqrt{10}$, $BC = \sqrt{9+9} = 3\sqrt{2}$, $CD = \sqrt{9+1} = \sqrt{10}$, $DA = \sqrt{9+9} = 3\sqrt{2}$. Diagonals $AC = 2$, $BD = \sqrt{36+16} = \sqrt{52}$. Opposite sides equal but diagonals unequal — $ABCD$ is a parallelogram.

Q7. Find the point on the X-axis which is equidistant from $(2,-5)$ and $(-2,9)$.

Answer: Any point on the X-axis is $(x,0)$. Setting distances equal:

$$(x-2)^2 + 25 = (x+2)^2 + 81$$

$$-4x + 4 + 25 = 4x + 4 + 81 \Rightarrow -8x = 56 \Rightarrow x = -7$$

The required point is $(-7, 0)$.

Q8. Find the values of $y$ for which the distance between the points $P(2,-3)$ and $Q(10,y)$ is $10$ units.

Answer:

$$\sqrt{64 + (y+3)^2} = 10 \Rightarrow (y+3)^2 = 36 \Rightarrow y+3 = \pm 6$$

$$y = 3 \text{ or } y = -9$$

Q9. If $Q(0,1)$ is equidistant from $P(5,-3)$ and $R(x,6)$, find the values of $x$. Also find the distances $QR$ and $PR$.

Answer: $QP = QR$ gives

$$25 + 16 = x^2 + 25 \Rightarrow x^2 = 16 \Rightarrow x = \pm 4$$

For $x=4$: $QR = \sqrt{16+25} = \sqrt{41}$, $PR = \sqrt{1+81} = \sqrt{82}$. For $x=-4$: $QR = \sqrt{41}$, $PR = \sqrt{81+81} = 9\sqrt{2}$.

Q10. Find a relation between $x$ and $y$ such that the point $(x,y)$ is equidistant from the points $(3,6)$ and $(-3,4)$.

Answer: Setting squared distances equal:

$$(x-3)^2 + (y-6)^2 = (x+3)^2 + (y-4)^2$$

Expanding and simplifying: $-6x – 12y + 45 = 6x – 8y + 25$, giving $\boxed{3x + 2y = 10}$ as the required relation.

Exercise 7.2 — Section Formula

Q1. Find the coordinates of the point which divides the join of $(-1,7)$ and $(4,-3)$ in the ratio $2:3$.

Answer: Using $\left(\frac{m x_2+n x_1}{m+n},\ \frac{m y_2+n y_1}{m+n}\right)$ with $m=2$, $n=3$:

$$x = \frac{2(4)+3(-1)}{5} = \frac{5}{5} = 1,\quad y = \frac{2(-3)+3(7)}{5} = \frac{15}{5} = 3$$

Required point: $(1, 3)$.

Q2. Find the coordinates of the points of trisection of the line segment joining $(4,-1)$ and $(-2,-3)$.

Answer: Trisection points $P$ and $Q$ divide $AB$ in ratios $1:2$ and $2:1$.

$P$: $\left(\dfrac{1(-2)+2(4)}{3}, \dfrac{1(-3)+2(-1)}{3}\right) = (2, -\tfrac{5}{3})$.

$Q$: $\left(\dfrac{2(-2)+1(4)}{3}, \dfrac{2(-3)+1(-1)}{3}\right) = (0, -\tfrac{7}{3})$.

Q3. To conduct sports day activities, in your rectangular school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD. Niharika runs $\frac{1}{4}$th the distance AD on the 2nd line and posts a green flag. Preet runs $\frac{1}{5}$th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?

Answer: Green flag: $G(2, 25)$; Red flag: $R(8, 20)$.

$$GR = \sqrt{(8-2)^2 + (20-25)^2} = \sqrt{36 + 25} = \sqrt{61} \text{ m}$$

Midpoint (blue flag): $\left(\dfrac{2+8}{2}, \dfrac{25+20}{2}\right) = (5, 22.5)$. So Rashmi posts the blue flag on the 5th line at a distance of $22.5$ m from $A$.

Q4. Find the ratio in which the line segment joining $(-3,10)$ and $(6,-8)$ is divided by $(-1,6)$.

Answer: Let the ratio be $k:1$. Then:

$$-1 = \frac{6k – 3}{k+1} \Rightarrow -k-1 = 6k-3 \Rightarrow 7k = 2 \Rightarrow k = \tfrac{2}{7}$$

Required ratio: $\boxed{2:7}$ (internal division).

Q5. Find the ratio in which the line segment joining $A(1,-5)$ and $B(-4,5)$ is divided by the X-axis. Also find the coordinates of the point of division.

Answer: The point on the X-axis has $y=0$. Let ratio be $k:1$:

$$0 = \frac{5k – 5}{k+1} \Rightarrow k = 1$$

So ratio is $1:1$. The point of division is $\left(\dfrac{-4+1}{2}, 0\right) = \left(-\tfrac{3}{2}, 0\right)$.

Q6. If $(1,2),(4,y),(x,6)$ and $(3,5)$ are the vertices of a parallelogram taken in order, find $x$ and $y$.

Answer: In a parallelogram, diagonals bisect each other. Midpoint of $AC$ = Midpoint of $BD$:

$$\left(\tfrac{1+x}{2}, \tfrac{2+6}{2}\right) = \left(\tfrac{4+3}{2}, \tfrac{y+5}{2}\right)$$

So $1+x = 7 \Rightarrow x = 6$ and $8 = y+5 \Rightarrow y = 3$.

Q7. Find the coordinates of point $A$, where $AB$ is the diameter of a circle whose centre is $(2,-3)$ and $B$ is $(1,4)$.

Answer: Centre is the midpoint of diameter:

$$\left(\tfrac{x+1}{2}, \tfrac{y+4}{2}\right) = (2, -3) \Rightarrow x = 3,\ y = -10$$

Therefore $A = (3, -10)$.

Q8. If $A$ and $B$ are $(-2,-2)$ and $(2,-4)$ respectively, find the coordinates of $P$ such that $AP = \frac{3}{7}AB$ and $P$ lies on the line segment $AB$.

Answer: $AP:PB = 3:4$. Using section formula with $m=3$, $n=4$:

$$P = \left(\frac{3(2)+4(-2)}{7}, \frac{3(-4)+4(-2)}{7}\right) = \left(-\tfrac{2}{7}, -\tfrac{20}{7}\right)$$

Q9. Find the coordinates of the points which divide the line segment joining $A(-2,2)$ and $B(2,8)$ into four equal parts.

Answer: Three points $P_1, P_2, P_3$ divide $AB$ in ratios $1:3$, $1:1$ and $3:1$.

$P_1 = \left(\dfrac{-2+2(-2)+1(2)}{4 ?}\right)$ — using internal section directly:

$P_1 = \left(\tfrac{1(2)+3(-2)}{4}, \tfrac{1(8)+3(2)}{4}\right) = (-1, \tfrac{14}{4}) = (-1, 3.5)$.

$P_2$ (midpoint) $= (0, 5)$.

$P_3 = \left(\tfrac{3(2)+1(-2)}{4}, \tfrac{3(8)+1(2)}{4}\right) = (1, 6.5)$.

Q10. Find the area of a rhombus if its vertices are $(3,0),(4,5),(-1,4)$ and $(-2,-1)$ taken in order.

Answer: Area of rhombus $= \tfrac{1}{2}\,|d_1 \times d_2|$ where $d_1, d_2$ are the diagonals.

$d_1$ = distance between $(3,0)$ and $(-1,4)$ = $\sqrt{16+16} = 4\sqrt{2}$.

$d_2$ = distance between $(4,5)$ and $(-2,-1)$ = $\sqrt{36+36} = 6\sqrt{2}$.

$$\text{Area} = \tfrac{1}{2}\times 4\sqrt{2}\times 6\sqrt{2} = \tfrac{1}{2}\times 48 = 24 \text{ sq units}$$

Exercise 7.3 — Area of Triangles

Q1. Find the area of the triangle whose vertices are:

(i) $(2,3),(-1,0),(2,-4)$

Answer: Using $\text{Area} = \tfrac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$:

$$=\tfrac{1}{2}|2(0-(-4)) + (-1)(-4-3) + 2(3-0)|$$

$$=\tfrac{1}{2}|8+7+6| = \tfrac{21}{2} = 10.5 \text{ sq units}$$

(ii) $(-5,-1),(3,-5),(5,2)$

Answer:

$$=\tfrac{1}{2}|-5(-5-2) + 3(2+1) + 5(-1+5)|$$

$$=\tfrac{1}{2}|35 + 9 + 20| = 32 \text{ sq units}$$

Q2. In each of the following find the value of ‘k’, for which the points are collinear:

(i) $(7,-2),(5,1),(3,k)$

Answer: Collinearity ⇒ Area = 0:

$$7(1-k) + 5(k+2) + 3(-2-1) = 0$$

$$7 – 7k + 5k + 10 – 9 = 0 \Rightarrow -2k + 8 = 0 \Rightarrow k = 4$$

(ii) $(8,1),(k,-4),(2,-5)$

Answer:

$$8(-4+5) + k(-5-1) + 2(1+4) = 0$$

$$8 – 6k + 10 = 0 \Rightarrow k = 3$$

Q3. Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are $(0,-1),(2,1)$ and $(0,3)$. Find the ratio of this area to the area of the given triangle.

Answer: Midpoints: $D(1,0),\ E(1,2),\ F(0,1)$.

Area of $\triangle DEF = \tfrac{1}{2}|1(2-1) + 1(1-0) + 0(0-2)| = \tfrac{1}{2}|1+1| = 1$ sq unit.

Area of original triangle $= \tfrac{1}{2}|0(1-3) + 2(3+1) + 0(-1-1)| = 4$ sq units.

Ratio $= 1:4$.

Q4. Find the area of the quadrilateral whose vertices, taken in order, are $(-4,-2),(-3,-5),(3,-2)$ and $(2,3)$.

Answer: Split along diagonal $AC$ into $\triangle ABC$ and $\triangle ACD$.

$\triangle ABC$ with $A(-4,-2), B(-3,-5), C(3,-2)$:

$=\tfrac{1}{2}|-4(-5+2) + (-3)(-2+2) + 3(-2+5)| = \tfrac{1}{2}|12+0+9| = 10.5$.

$\triangle ACD$ with $A(-4,-2), C(3,-2), D(2,3)$:

$=\tfrac{1}{2}|-4(-2-3) + 3(3+2) + 2(-2+2)| = \tfrac{1}{2}|20+15+0| = 17.5$.

Total area $= 10.5 + 17.5 = 28$ sq units.

Q5. You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for $\triangle ABC$ whose vertices are $A(4,-6),B(3,-2)$ and $C(5,2)$.

Answer: Midpoint $D$ of $BC$ = $(4, 0)$.

Area of $\triangle ABD = \tfrac{1}{2}|4(-2-0)+3(0+6)+4(-6+2)| = \tfrac{1}{2}|-8+18-16| = 3$ sq units.

Area of $\triangle ACD = \tfrac{1}{2}|4(2-0)+5(0+6)+4(-6-2)| = \tfrac{1}{2}|8+30-32| = 3$ sq units.

Both areas equal — median $AD$ divides $\triangle ABC$ into two triangles of equal area. Verified.

Exercise 7.4 (Optional)

Q1. Determine the ratio in which the line $2x+y-4=0$ divides the line segment joining $A(2,-2)$ and $B(3,7)$.

Answer: Let ratio be $k:1$. Point of division:

$$\left(\tfrac{3k+2}{k+1},\ \tfrac{7k-2}{k+1}\right)$$

This lies on $2x+y-4=0$:

$$2\cdot\tfrac{3k+2}{k+1} + \tfrac{7k-2}{k+1} – 4 = 0$$

$$6k+4+7k-2-4(k+1) = 0 \Rightarrow 9k – 2 = 0 \Rightarrow k = \tfrac{2}{9}$$

Required ratio: $\boxed{2:9}$.

Q2. Find a relation between $x$ and $y$ if the points $(x,y),(1,2)$ and $(7,0)$ are collinear.

Answer: Area = 0:

$$x(2-0)+1(0-y)+7(y-2) = 0 \Rightarrow 2x – y + 7y – 14 = 0$$

$$2x + 6y – 14 = 0 \Rightarrow x + 3y = 7$$

Q3. Find the centre of a circle passing through the points $(6,-6),(3,-7)$ and $(3,3)$.

Answer: Let centre $= (h,k)$. Equate distances:

$(h-6)^2+(k+6)^2 = (h-3)^2+(k+7)^2$ ⇒ $-6h – 2k – 14 = 0$ ⇒ $3h+k=-7$ … (i)

$(h-3)^2+(k+7)^2 = (h-3)^2+(k-3)^2$ ⇒ $20k + 40 = 0$ ⇒ $k = -2$.

From (i): $3h – 2 = -7 \Rightarrow h = -\tfrac{5}{3}$. Wait, recompute: $3h + (-2) = -7 \Rightarrow 3h = -5 \Rightarrow h = -\tfrac{5}{3}$. Re-checking the standard answer with the textbook gives centre $(3, -3)$. Let us recompute carefully:

$(h-6)^2+(k+6)^2 = (h-3)^2+(k+7)^2$ expands to $-6h – 2k = 14$, i.e. $3h+k = -7$. Hmm.

$(h-3)^2 + (k+7)^2 = (h-3)^2 + (k-3)^2$ gives $20k = -40$, so $k = -2$. Then $h = -5/3$ — but recheck: cross-multiplying NCERT solution gives centre $(3,-3)$. Using $k=-2$ and rechecking the first equation: $(h-6)^2+( -2+6)^2 = (h-3)^2+(-2+7)^2 \Rightarrow (h-6)^2 + 16 = (h-3)^2 + 25 \Rightarrow h^2-12h+36+16 = h^2-6h+9+25 \Rightarrow -6h + 18 = 0 \Rightarrow h = 3$. So centre $= (3, -2)$.

Centre: $(3, -2)$. (The earlier $3h+k=-7$ contained a sign slip; the correct simplification yields $h=3$.)

Q4. The two opposite vertices of a square are $(-1,2)$ and $(3,2)$. Find the coordinates of the other two vertices.

Answer: Let the other vertices be $B(x,y)$ and $D$. Diagonals of a square are equal and bisect at right angles; so $B$ lies on the perpendicular bisector of $AC$ at the same distance from the centre as $A$. Centre = midpoint of $AC$ = $(1,2)$. Half-diagonal = $2$. Perpendicular bisector of $AC$ is the vertical line $x = 1$. Hence $B(1, 2+2) = (1, 4)$ and $D(1, 2-2) = (1, 0)$.

Q5. The Class X students of a secondary school have been allotted a rectangular plot of land for their gardening activity. Sapling of Gulmohar are planted on the boundary at a distance of 1m from each other. There is a triangular grassy lawn in the plot as shown in the figure. The students are to sow seeds of flowering plants on the remaining area of the plot.

(i) Taking A as origin, find the coordinates of the vertices of the triangle.

Answer: $P(4, 6),\ Q(3, 2),\ R(6, 5)$.

(ii) What will be the coordinates of the vertices of $\triangle PQR$ if C is the origin?

Answer: $P(-12, -2),\ Q(-13, -6),\ R(-10, -3)$.

Areas: In both cases, area of $\triangle PQR$ = $\tfrac{1}{2}|4(2-5)+3(5-6)+6(6-2)| = \tfrac{1}{2}|-12-3+24| = 4.5$ sq units. Same in both systems — area is invariant under translation.

Q6. The vertices of a $\triangle ABC$ are $A(4,6),B(1,5)$ and $C(7,2)$. A line is drawn to intersect sides $AB$ and $AC$ at $D$ and $E$ respectively, such that $\frac{AD}{AB}=\frac{AE}{AC}=\frac{1}{4}$. Calculate the area of $\triangle ADE$ and compare it with the area of $\triangle ABC$.

Answer: $D$ divides $AB$ in $1:3$, so $D = \left(\tfrac{1+12}{4}, \tfrac{5+18}{4}\right) = \left(\tfrac{13}{4}, \tfrac{23}{4}\right)$.

$E$ divides $AC$ in $1:3$, so $E = \left(\tfrac{7+12}{4}, \tfrac{2+18}{4}\right) = \left(\tfrac{19}{4}, 5\right)$.

Area of $\triangle ADE = \tfrac{1}{2}\left|4\left(\tfrac{23}{4}-5\right) + \tfrac{13}{4}(5-6) + \tfrac{19}{4}\left(6-\tfrac{23}{4}\right)\right| = \tfrac{15}{32}$ sq units.

Area of $\triangle ABC = \tfrac{1}{2}|4(5-2)+1(2-6)+7(6-5)| = \tfrac{1}{2}|12-4+7| = \tfrac{15}{2}$ sq units.

Ratio $= \tfrac{15/32}{15/2} = \tfrac{1}{16}$. So $\triangle ADE : \triangle ABC = 1:16$.

Q7. Let $A(4,2),B(6,5)$ and $C(1,4)$ be the vertices of $\triangle ABC$.

(i) The median from $A$ meets $BC$ at $D$. Find the coordinates of the point $D$.

Answer: $D$ = midpoint of $BC$ = $\left(\tfrac{6+1}{2}, \tfrac{5+4}{2}\right) = \left(\tfrac{7}{2}, \tfrac{9}{2}\right)$.

(ii) Find the coordinates of the point $P$ on $AD$ such that $AP:PD=2:1$.

Answer: $P = \left(\tfrac{2(7/2)+1(4)}{3}, \tfrac{2(9/2)+1(2)}{3}\right) = \left(\tfrac{11}{3}, \tfrac{11}{3}\right)$.

(iii) Find the coordinates of points $Q$ and $R$ on medians $BE$ and $CF$ respectively, such that $BQ:QE=2:1$ and $CR:RF=2:1$.

Answer: $E$ = midpoint of $AC$ = $(2.5, 3)$. $Q$ on $BE$ with $BQ:QE=2:1$: $Q=\left(\tfrac{11}{3},\tfrac{11}{3}\right)$. Similarly $R=\left(\tfrac{11}{3},\tfrac{11}{3}\right)$.

(iv) What do you observe?

Answer: $P=Q=R=\left(\tfrac{11}{3},\tfrac{11}{3}\right)$ — all three medians pass through the same point, the centroid, which divides each median in $2:1$.

(v) If $A(x_1,y_1),B(x_2,y_2)$ and $C(x_3,y_3)$ are the vertices of $\triangle ABC$, find the coordinates of the centroid of the triangle.

Answer: Centroid $G = \left(\dfrac{x_1+x_2+x_3}{3},\ \dfrac{y_1+y_2+y_3}{3}\right)$.

Q8. $ABCD$ is a rectangle formed by the points $A(-1,-1),B(-1,4),C(5,4)$ and $D(5,-1)$. $P,Q,R,S$ are the mid-points of $AB,BC,CD,DA$ respectively. Is the quadrilateral $PQRS$ a square, rectangle or rhombus? Justify.

Answer: $P(-1, 1.5),\ Q(2, 4),\ R(5, 1.5),\ S(2,-1)$.

$PQ = \sqrt{9 + 6.25} = \sqrt{15.25}$, $QR = \sqrt{9+6.25} = \sqrt{15.25}$, $RS = \sqrt{9+6.25} = \sqrt{15.25}$, $SP = \sqrt{9+6.25} = \sqrt{15.25}$. All sides equal.

Diagonals: $PR = 6$, $QS = 5$. Diagonals unequal — so $PQRS$ is a rhombus (not a square).

Worked Examples (between exercises)

Worked Example 1. Show that the points $A(1,7)$, $B(4,2)$, $C(-1,-1)$ and $D(-4,4)$ form a square.

Solution. All four sides should be equal and the diagonals should also be equal.

$AB = \sqrt{(4-1)^2+(2-7)^2} = \sqrt{9+25} = \sqrt{34}$.

$BC = \sqrt{(-1-4)^2+(-1-2)^2} = \sqrt{25+9} = \sqrt{34}$.

$CD = \sqrt{(-4+1)^2+(4+1)^2} = \sqrt{9+25} = \sqrt{34}$.

$DA = \sqrt{(1+4)^2+(7-4)^2} = \sqrt{25+9} = \sqrt{34}$.

Diagonals: $AC = \sqrt{(-1-1)^2+(-1-7)^2} = \sqrt{4+64} = \sqrt{68}$ and $BD = \sqrt{(-4-4)^2+(4-2)^2} = \sqrt{64+4} = \sqrt{68}$. All four sides equal, both diagonals equal — $ABCD$ is a square.

Worked Example 2. Find the coordinates of the point which divides the segment joining $A(1,-3)$ and $B(4,5)$ in the ratio $3:5$ internally.

Solution. With $m = 3$, $n = 5$:

$$x = \frac{3(4) + 5(1)}{3+5} = \frac{17}{8},\quad y = \frac{3(5) + 5(-3)}{8} = 0$$

Required point: $\left(\tfrac{17}{8},\, 0\right)$. Notice that this point lies on the X-axis.

Worked Example 3. Show that the points $(2,-2),(-2,1),(5,2)$ are vertices of a right-angled triangle. Also find the area.

Solution. Let $A(2,-2)$, $B(-2,1)$, $C(5,2)$.

$AB^2 = 16+9 = 25$, $BC^2 = 49+1 = 50$, $AC^2 = 9+16 = 25$.

Since $AB^2 + AC^2 = 25 + 25 = 50 = BC^2$, the triangle is right-angled at $A$ (Pythagoras converse).

Area $= \tfrac{1}{2}\cdot AB\cdot AC = \tfrac{1}{2}\cdot 5\cdot 5 = 12.5$ sq units.

Worked Example 4. Find the centroid of the triangle with vertices $(2,3),(-4,5),(5,-2)$.

Solution.

$$G = \left(\frac{2-4+5}{3},\ \frac{3+5-2}{3}\right) = (1, 2)$$

Worked Example 5. If the point $(x,y)$ is equidistant from $(2,1)$ and $(1,-2)$, find a relation between $x$ and $y$.

Solution. $(x-2)^2 + (y-1)^2 = (x-1)^2 + (y+2)^2$. Expanding:

$x^2 -4x+4 +y^2-2y+1 = x^2-2x+1+y^2+4y+4$.

Cancelling and simplifying: $-4x – 2y + 5 = -2x + 4y + 5 \Rightarrow -2x – 6y = 0 \Rightarrow x + 3y = 0$.

Worked Example 6. The line segment joining $(2,-2)$ and $(-7,4)$ is to be divided into three equal parts. Find the points of division.

Solution. Trisection ratios are $1:2$ and $2:1$.

$P_1 = \left(\dfrac{1(-7)+2(2)}{3}, \dfrac{1(4)+2(-2)}{3}\right) = \left(-1, 0\right)$.

$P_2 = \left(\dfrac{2(-7)+1(2)}{3}, \dfrac{2(4)+1(-2)}{3}\right) = \left(-4, 2\right)$.

Worked Example 7. Show that the points $(3,0),(-1,-1)$ and $(2,3)$ are vertices of an isosceles triangle.

Solution. $AB = \sqrt{16+1} = \sqrt{17}$; $BC = \sqrt{9+16} = 5$; $AC = \sqrt{1+9} = \sqrt{10}$. Since $AB \neq BC \neq AC$ — let us re-check: $AB=\sqrt{17}\approx 4.12$, $BC=5$, $AC=\sqrt{10}\approx 3.16$. Hmm — these are unequal, so the points actually do not form an isosceles triangle. Substitute $A(3,0),\ B(-1,-1),\ C(2,3)$ once more carefully: $AC = \sqrt{(2-3)^2+(3-0)^2} = \sqrt{1+9} = \sqrt{10}$. So $AB = \sqrt{17}$, $BC = 5 = \sqrt{25}$, $AC = \sqrt{10}$ — three different lengths, scalene triangle. (This shows the importance of computing carefully before declaring a result!)

Worked Example 8. Find the value of $k$ if the points $A(2,3), B(4,k), C(6,-3)$ are collinear.

Solution. Set area $=0$:

$$2(k+3) + 4(-3-3) + 6(3-k) = 0$$

$$2k + 6 – 24 + 18 – 6k = 0 \Rightarrow -4k = 0 \Rightarrow k = 0$$

Worked Example 9. Find the area of the quadrilateral $ABCD$ with $A(1,1)$, $B(7,-3)$, $C(12,2)$, $D(7,21)$.

Solution. Diagonal $AC$ splits it into $\triangle ABC$ and $\triangle ACD$.

$\triangle ABC = \tfrac{1}{2}|1(-3-2) + 7(2-1) + 12(1+3)| = \tfrac{1}{2}|-5+7+48| = 25$ sq units.

$\triangle ACD = \tfrac{1}{2}|1(2-21) + 12(21-1) + 7(1-2)| = \tfrac{1}{2}|-19+240-7| = 107$ sq units.

Area of quadrilateral $= 25 + 107 = 132$ sq units.

Worked Example 10. If $A(-2,-1)$, $B(a,0)$, $C(4,b)$ and $D(1,2)$ are vertices of a parallelogram, find $a$ and $b$.

Solution. Diagonals of a parallelogram bisect each other:

$$\left(\tfrac{-2+4}{2},\tfrac{-1+b}{2}\right) = \left(\tfrac{a+1}{2},\tfrac{0+2}{2}\right)$$

$1 = \tfrac{a+1}{2} \Rightarrow a = 1$ and $\tfrac{-1+b}{2} = 1 \Rightarrow b = 3$.

Additional Practice Questions

Q1. Find the distance of the point $(-3, 4)$ from the origin.

Answer: $OP = \sqrt{(-3)^2 + 4^2} = \sqrt{9+16} = 5$ units.

Q2. Find the coordinates of the midpoint of the line segment joining $(-2, 7)$ and $(6, -3)$.

Answer: $M = \left(\tfrac{-2+6}{2}, \tfrac{7-3}{2}\right) = (2, 2)$.

Q3. Show that the points $(1,7),(4,2),(-1,-1)$ and $(-4,4)$ are the vertices of a square.

Answer: All four sides equal $\sqrt{34}$ and both diagonals equal $\sqrt{68}$ — it is a square.

Q4. Find the area of the triangle whose vertices are $(1,-1),(-4,6)$ and $(-3,-5)$.

Answer:

$$\text{Area} = \tfrac{1}{2}|1(6+5)+(-4)(-5+1)+(-3)(-1-6)| = \tfrac{1}{2}|11+16+21| = 24 \text{ sq units}$$

Q5. The point which divides the line segment joining $(2,-5)$ and $(5,2)$ in the ratio $2:3$ internally lies in which quadrant?

Answer: Point $= \left(\tfrac{2(5)+3(2)}{5}, \tfrac{2(2)+3(-5)}{5}\right) = \left(\tfrac{16}{5}, -\tfrac{11}{5}\right)$. Lies in Quadrant IV.

Q6. Find the perimeter of the triangle whose vertices are $(0,4),(0,0)$ and $(3,0)$.

Answer: Sides: $4$, $3$, $\sqrt{9+16}=5$. Perimeter $= 12$ units.

Q7. If the distance between the points $(4, p)$ and $(1, 0)$ is $5$, find the value of $p$.

Answer: $\sqrt{9+p^2} = 5 \Rightarrow p^2 = 16 \Rightarrow p = \pm 4$.

Q8. Find the area of the triangle with vertices $(2,3),(-1,0)$ and $(2,-4)$ — verify using the determinant arrangement.

Answer: $= \tfrac{1}{2}|2(0+4)+(-1)(-4-3)+2(3-0)| = \tfrac{1}{2}|8+7+6| = 10.5$ sq units.

Q9. Find the ratio in which the Y-axis divides the line segment joining $(5,-6)$ and $(-1,-4)$.

Answer: On the Y-axis $x=0$. Let ratio $k:1$:

$$0 = \tfrac{-k+5}{k+1} \Rightarrow k = 5$$

Ratio $= 5:1$. Point $= \left(0, \tfrac{-20-6}{6}\right) = \left(0, -\tfrac{13}{3}\right)$.

Q10. The line segment joining $A(6,3)$ and $B(-1,-4)$ is divided internally in the ratio $2:1$ at $P$. Find $P$.

Answer: $P = \left(\tfrac{2(-1)+1(6)}{3}, \tfrac{2(-4)+1(3)}{3}\right) = \left(\tfrac{4}{3}, -\tfrac{5}{3}\right)$.

Q11. Show that the points $(0,-1),(2,1),(0,3),(-2,1)$ form a square.

Answer: All four sides equal $\sqrt{4+4}=2\sqrt{2}$. Diagonals: $\sqrt{0+16}=4$ and $\sqrt{16+0}=4$ — both equal. Hence a square.

Q12. Find the value of $y$ for which the points $(1,4),(3,y),(-3,16)$ are collinear.

Answer: Set area $=0$: $1(y-16) + 3(16-4) + (-3)(4-y) = 0 \Rightarrow y -16 +36 – 12 +3y = 0 \Rightarrow 4y +8 = 0 \Rightarrow y = -2$.

Q13. Find the centroid of the triangle whose vertices are $(-1, 0), (5,-2), (8,2)$.

Answer: $G = \left(\tfrac{-1+5+8}{3}, \tfrac{0-2+2}{3}\right) = (4, 0)$.

Q14. The coordinates of one end of a diameter of a circle are $(2,3)$ and the centre is $(-2,5)$. Find the other end.

Answer: Let other end be $(x,y)$. Centre = midpoint:

$\left(\tfrac{2+x}{2},\tfrac{3+y}{2}\right) = (-2,5) \Rightarrow x = -6, y = 7$. So $(-6,7)$.

Q15. Find the points of trisection of the line segment joining the points $(3,-2)$ and $(-3,-4)$.

Answer: $P_1 = \left(\tfrac{1(-3)+2(3)}{3}, \tfrac{1(-4)+2(-2)}{3}\right) = \left(1, -\tfrac{8}{3}\right)$.

$P_2 = \left(\tfrac{2(-3)+1(3)}{3}, \tfrac{2(-4)+1(-2)}{3}\right) = \left(-1, -\tfrac{10}{3}\right)$.

Q16. The points $A(2,9),B(a,5),C(5,5)$ are vertices of a right triangle right-angled at $B$. Find $a$.

Answer: $AB^2 + BC^2 = AC^2$: $(2-a)^2 + 16 + (a-5)^2 + 0 = 9 + 16$. Expanding: $a^2-4a+4+16+a^2-10a+25 = 25 \Rightarrow 2a^2 -14a + 20 = 0 \Rightarrow a^2 -7a + 10 = 0 \Rightarrow a = 2$ or $a = 5$.

Q17. If the midpoint of the segment joining $A(2a,4)$ and $B(-2,3b)$ is $(1,2a+1)$, find $a$ and $b$.

Answer: $\tfrac{2a-2}{2} = 1 \Rightarrow a = 2$. $\tfrac{4+3b}{2} = 2a+1 = 5 \Rightarrow b = 2$.

Q18. Find the area of the triangle whose vertices are $(0,0),(0,5),(7,0)$.

Answer: This is a right triangle with legs 5 and 7. Area $= \tfrac{1}{2}\times 5\times 7 = 17.5$ sq units.

Q19. Determine if the points $(1,-1),(5,2),(9,5)$ are collinear.

Answer: Area $= \tfrac{1}{2}|1(2-5)+5(5+1)+9(-1-2)| = \tfrac{1}{2}|-3+30-27| = 0$. Yes — they are collinear.

Q20. Find the ratio in which the X-axis divides the segment joining $(2,-3)$ and $(5,6)$. Hence find the point of division.

Answer: Let ratio be $k:1$, $y=0$:

$0 = \tfrac{6k-3}{k+1} \Rightarrow k = \tfrac{1}{2}$. Ratio $= 1:2$. Point $= \left(\tfrac{1(5)+2(2)}{3},0\right) = (3,0)$.

Q21. The point $P$ divides the segment $A(-3,-4)$ and $B(-8,7)$ in the ratio $5:8$. Find $P$.

Answer: $P = \left(\tfrac{5(-8)+8(-3)}{13}, \tfrac{5(7)+8(-4)}{13}\right) = \left(\tfrac{-64}{13}, \tfrac{3}{13}\right)$.

Q22. Find the area of the triangle formed by joining the points $(-1.5, 3),(6,-2),(-3,4)$.

Answer:

$$=\tfrac{1}{2}|(-1.5)(-2-4)+6(4-3)+(-3)(3+2)|$$

$$=\tfrac{1}{2}|9+6-15| = 0 \text{ sq units}$$

Area is zero — the three points are collinear, so no triangle is formed.

Q23. The vertices of a triangle are $A(3,4),B(-4,3),C(7,-1)$. Find the length of the median through $A$.

Answer: Midpoint of $BC$ is $D = \left(\tfrac{3}{2},1\right)$. Median $AD = \sqrt{(3-1.5)^2+(4-1)^2} = \sqrt{2.25+9} = \sqrt{11.25} = \tfrac{3\sqrt{5}}{2}$ units.

Q24. Find the value of $p$ for which the distance between $(1,3)$ and $(p, 7)$ is $5$.

Answer: $(p-1)^2 + 16 = 25 \Rightarrow (p-1)^2 = 9 \Rightarrow p-1 = \pm 3 \Rightarrow p = 4 \text{ or } -2$.

Q25. The points $(0,5),(0,-9)$ and $(3,6)$ are vertices of which type of triangle?

Answer: $AB = \sqrt{0+196} = 14$; $BC = \sqrt{9+225} = \sqrt{234}$; $AC = \sqrt{9+1} = \sqrt{10}$. Three unequal sides — scalene triangle.

Higher Order Thinking Skills (HOTS) Problems

HOTS 1. If $A(2,2),B(-2,-2)$ and $C(-2\sqrt{3},2\sqrt{3})$ are the vertices of a triangle, prove that the triangle is equilateral.

Answer: $AB = \sqrt{16+16} = 4\sqrt{2}$. $BC = \sqrt{(-2+2\sqrt{3})^2 + (-2-2\sqrt{3})^2}$. Let us expand: $(-2+2\sqrt{3})^2 = 4 – 8\sqrt{3} + 12 = 16 – 8\sqrt{3}$ and $(-2-2\sqrt{3})^2 = 4 + 8\sqrt{3} + 12 = 16 + 8\sqrt{3}$. Sum $= 32$. So $BC = \sqrt{32} = 4\sqrt{2}$. Similarly $AC = \sqrt{(2+2\sqrt{3})^2 + (2-2\sqrt{3})^2} = \sqrt{(16+8\sqrt 3)+(16-8\sqrt 3)} = \sqrt{32} = 4\sqrt{2}$. All three sides equal — equilateral.

HOTS 2. The two opposite vertices of a square are $(0,0)$ and $(4,4)$. Find the coordinates of the other two vertices.

Answer: Let the missing vertices be $B(x,y)$ and $D(x’,y’)$. Centre of the square is the midpoint of either diagonal, which equals $(2,2)$. The other diagonal has the same midpoint and equal length $4\sqrt{2}$, but is perpendicular to $AC$. Since $AC$ has slope $1$, the perpendicular through $(2,2)$ has slope $-1$ and equation $y = -x + 4$. Length from centre to vertex along this diagonal $= 2\sqrt{2}$. Parametrise along $(\cos\theta,\sin\theta) = (\tfrac{1}{\sqrt 2},-\tfrac{1}{\sqrt 2})$: vertices at $(2+2,2-2) = (4,0)$ and $(2-2,2+2) = (0,4)$. So $B(4,0)$ and $D(0,4)$.

HOTS 3. If three vertices of a parallelogram, taken in order, are $(-1,0),(3,1),(2,2)$, find the fourth vertex.

Answer: Diagonals bisect each other. Midpoint of diagonal $AC$ = midpoint of diagonal $BD$. With $A(-1,0)$, $B(3,1)$, $C(2,2)$, $D(x,y)$:

$\left(\tfrac{-1+2}{2},\tfrac{0+2}{2}\right) = \left(\tfrac{3+x}{2},\tfrac{1+y}{2}\right) \Rightarrow x = -2,\ y = 1$. Fourth vertex $D(-2,1)$.

HOTS 4. Find the value of $k$ so that the triangle with vertices $(k,4),(2,-6),(5,4)$ has area $35$ sq units.

Answer: $\tfrac{1}{2}|k(-6-4)+2(4-4)+5(4+6)| = 35 \Rightarrow |{-10k+50}| = 70 \Rightarrow -10k+50 = \pm 70 \Rightarrow k = -2$ or $k = 12$.

HOTS 5. Three vertices of a triangle $ABC$ are $A(1,2)$, $B(-3,2)$ and $C(3,4)$. Find the equation of the line passing through the midpoint of $BC$ and the vertex $A$.

Answer: Midpoint of $BC$ is $D(0,3)$. Slope of $AD = \tfrac{3-2}{0-1} = -1$. Equation: $y – 2 = -1(x-1) \Rightarrow x + y = 3$.

HOTS 6. If the points $A(1,2),B(0,0)$ and $C(a,b)$ are collinear, find the relation between $a$ and $b$. Also find the distance $AC$ when $a=3$.

Answer: Collinearity: $1(0-b)+0(b-2)+a(2-0) = 0 \Rightarrow 2a = b$. So $b=2a$. For $a=3$, $b=6$. $AC = \sqrt{(3-1)^2+(6-2)^2} = \sqrt{4+16} = 2\sqrt{5}$ units.

HOTS 7. The line segment joining $A(2,1)$ and $B(5,-8)$ is trisected at the points $P$ and $Q$ such that $P$ is nearer to $A$. If $P$ also lies on the line $2x-y+k=0$, find $k$.

Answer: $P$ divides $AB$ in $1:2$, so $P = \left(\tfrac{1(5)+2(2)}{3}, \tfrac{1(-8)+2(1)}{3}\right) = (3,-2)$. Substituting in $2x-y+k=0$: $6+2+k=0 \Rightarrow k = -8$.

HOTS 8. Find the coordinates of the points which trisect the line segment $PQ$ where $P(2,-2)$ and $Q(-7,4)$. Also verify that the midpoint of these two trisection points coincides with the midpoint of $PQ$.

Answer: $T_1(\text{at }1:2) = \left(\tfrac{-7+4}{3},\tfrac{4-4}{3}\right) = (-1,0)$. $T_2(\text{at }2:1) = \left(\tfrac{-14+2}{3},\tfrac{8-2}{3}\right) = (-4,2)$. Midpoint of $T_1 T_2 = (-2.5, 1)$. Midpoint of $PQ = (-2.5, 1)$. They coincide — verified.

HOTS 9. Show that the points $(a,a),(-a,-a),(-a\sqrt{3},a\sqrt{3})$ are vertices of an equilateral triangle. Also find its area.

Answer: $AB = \sqrt{4a^2+4a^2} = 2a\sqrt{2}$. $BC = \sqrt{(-a\sqrt 3 + a)^2+(a\sqrt 3 + a)^2} = \sqrt{(a^2 – 2a^2\sqrt 3 + 3a^2)+(3a^2+2a^2\sqrt 3 + a^2)} = \sqrt{8a^2} = 2a\sqrt{2}$. Similarly $AC = 2a\sqrt{2}$. Equilateral. Area $= \tfrac{\sqrt 3}{4}(2a\sqrt 2)^2 = \tfrac{\sqrt 3}{4}\cdot 8a^2 = 2\sqrt 3\, a^2$ sq units.

HOTS 10. The line joining $A(4,-5)$ and $B(4,5)$ is divided by the point $P$ in the ratio $1:2$ from $A$. Find $P$ and the distance $AP$.

Answer: $P = \left(\tfrac{1(4)+2(4)}{3},\tfrac{1(5)+2(-5)}{3}\right) = \left(4,-\tfrac{5}{3}\right)$. $AP = \sqrt{0 + \left(-\tfrac{5}{3}+5\right)^2} = \sqrt{\left(\tfrac{10}{3}\right)^2} = \tfrac{10}{3}$ units.

Real-life Application Problems

App 1. A field is laid out as a coordinate grid with sides 1 m apart. A blue flag is at $(3,4)$ and a red flag at $(7,7)$. How far apart (in metres) are the flags?

Answer: $d = \sqrt{16+9} = 5$ m.

App 2. Two cell-phone towers stand at coordinates $(2,3)$ km and $(11,12)$ km. Find the location of a midway repeater station.

Answer: Midpoint $= \left(\tfrac{13}{2},\tfrac{15}{2}\right) = (6.5, 7.5)$ km.

App 3. A rectangular plot $ABCD$ has $A(0,0)$, $B(40,0)$, $C(40,30)$, $D(0,30)$ (units in metres). Find the length of the diagonal $AC$ and the centre of the plot.

Answer: $AC = \sqrt{1600+900} = 50$ m. Centre = midpoint of $AC$ = $(20,15)$ m.

App 4. Three friends are standing at $A(1,1),B(5,1),C(3,4)$ on a coordinate grid. Where would they meet if each one walks to the centroid of the triangle they form?

Answer: $G = \left(\tfrac{1+5+3}{3},\tfrac{1+1+4}{3}\right) = (3,2)$.

App 5. A garden has a triangular flower-bed with vertices $A(2,1),B(5,4),C(8,1)$ (in metres). Find the area available for planting.

Answer: $\tfrac{1}{2}|2(4-1)+5(1-1)+8(1-4)| = \tfrac{1}{2}|6+0-24| = 9$ sq m.

App 6. A drone takes off from base $(0,0)$ and lands at $(60, 80)$ metres. What is the straight-line distance covered?

Answer: $\sqrt{3600+6400} = \sqrt{10000} = 100$ m.

App 7. A school has placed three water taps at $(2,3), (5,3)$ and $(3.5,7)$. To minimise pipe length, where should a junction box be located to be equidistant from the first two taps?

Answer: The locus of points equidistant from $(2,3)$ and $(5,3)$ is the perpendicular bisector — the vertical line $x = 3.5$. Choosing $y = 3$ gives the point closest to all three: $(3.5, 3)$.

App 8. A satellite ground station at $(0,0)$ tracks a satellite that moves from $(3,4)$ to $(-5,12)$ in some interval. Find the change in distance from the station between the two positions.

Answer: Initial distance $= 5$, final distance $= \sqrt{25+144} = 13$. Change $= 8$ units.

Multiple Choice Questions (MCQ)

1. The distance between the points $(0,0)$ and $(-8,6)$ is (a) 10 (b) 8 (c) 14 (d) 6.

Answer: $\sqrt{64+36} = \sqrt{100} = 10$. Option (a).

2. The distance of $P(2,3)$ from the X-axis is (a) 2 (b) -3 (c) -2 (d) 3.

Answer: Distance from X-axis = $|y|$ = 3. Option (d).

3. The distance between $(a,b)$ and $(-a,-b)$ is (a) $2\sqrt{a^2+b^2}$ (b) $\sqrt{a^2+b^2}$ (c) $a^2+b^2$ (d) $2(a+b)$.

Answer: $\sqrt{(2a)^2+(2b)^2} = 2\sqrt{a^2+b^2}$. Option (a).

4. The midpoint of the segment joining $(2,3)$ and $(4,7)$ is (a) $(3,5)$ (b) $(6,10)$ (c) $(2,4)$ (d) $(1,2)$.

Answer: $\left(\tfrac{2+4}{2},\tfrac{3+7}{2}\right) = (3,5)$. Option (a).

5. If the points $A(1,2),O(0,0),C(a,b)$ are collinear, then (a) $a=b$ (b) $a=2b$ (c) $2a=b$ (d) $a+b=0$.

Answer: Set area $=0$: $1(0-b)+0+a(2-0)=0 \Rightarrow -b+2a = 0 \Rightarrow 2a = b$. Option (c).

6. The point on the X-axis equidistant from $(2,-5)$ and $(-2,9)$ is (a) $(7,0)$ (b) $(-7,0)$ (c) $(0,7)$ (d) $(0,-7)$.

Answer: From Exercise 7.1 Q7 — $(-7,0)$. Option (b).

7. The coordinates of the centroid of triangle with vertices $(1,2),(3,-1),(5,5)$ are (a) $(3,2)$ (b) $(2,3)$ (c) $(3,3)$ (d) $(9,6)$.

Answer: $\left(\tfrac{1+3+5}{3},\tfrac{2-1+5}{3}\right) = (3,2)$. Option (a).

8. The point that divides the segment joining $(-1,3)$ and $(4,-7)$ in the ratio $3:4$ internally is (a) $\left(\tfrac{8}{7},-\tfrac{9}{7}\right)$ (b) $\left(\tfrac{8}{7},\tfrac{9}{7}\right)$ (c) $(2,-3)$ (d) $(-2,3)$.

Answer: $x = \tfrac{3(4)+4(-1)}{7} = \tfrac{8}{7}$, $y = \tfrac{3(-7)+4(3)}{7} = -\tfrac{9}{7}$. Option (a).

9. The area of the triangle with vertices $(0,0),(5,0),(0,4)$ is (a) 20 (b) 10 (c) 9 (d) 5.

Answer: Right triangle with legs 5 and 4. Area $= \tfrac{1}{2}(5)(4) = 10$. Option (b).

10. If the distance between $(x,2)$ and $(3,-6)$ is 10, then $x=$ (a) 3 only (b) -3 only (c) 9 or -3 (d) ±3.

Answer: $(x-3)^2 + 64 = 100 \Rightarrow (x-3)^2 = 36 \Rightarrow x = 9$ or $x = -3$. Option (c).

11. The point $(-3,5)$ lies in (a) Q I (b) Q II (c) Q III (d) Q IV.

Answer: $x<0$, $y>0$ — Quadrant II. Option (b).

12. The distance of the point $(0,-7)$ from the origin is (a) 0 (b) 7 (c) -7 (d) 49.

Answer: $\sqrt{0+49} = 7$. Option (b).

13. The midpoint of the segment joining $(-2,8)$ and $(-6,-4)$ is (a) $(-4,2)$ (b) $(2,-4)$ (c) $(4,2)$ (d) $(-4,-2)$.

Answer: $\left(\tfrac{-8}{2},\tfrac{4}{2}\right) = (-4,2)$. Option (a).

14. The Y-axis divides the segment joining $(-4,5)$ and $(3,-7)$ in the ratio (a) $4:3$ (b) $3:4$ (c) $5:7$ (d) $7:5$.

Answer: Let ratio $k:1$, $x=0$: $0 = \tfrac{3k-4}{k+1} \Rightarrow k = \tfrac{4}{3}$. Ratio $= 4:3$. Option (a).

15. The points $(1,5),(2,3),(-2,-11)$ are (a) collinear (b) form a right triangle (c) form an isosceles triangle (d) none of these.

Answer: By Exercise 7.1 Q3 they are not collinear; testing the others shows none — option (d).

Long Answer (Board-style) Questions

LA 1. State and prove the distance formula in coordinate geometry. Use it to find the distance between $A(7,5)$ and $B(2,5)$, and verify that the result equals the absolute difference of the abscissae.

Answer: The distance formula states: for two points $A(x_1,y_1)$ and $B(x_2,y_2)$ in the Cartesian plane, the distance between them is $AB = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.

Proof: Drop perpendiculars from $A$ and $B$ on the X-axis at $L$ and $M$. From $A$ draw a horizontal line meeting $BM$ at $N$. Then $AN = LM = x_2 – x_1$ and $BN = y_2 – y_1$. Triangle $ANB$ is right-angled at $N$, so by Pythagoras’ theorem:

$$AB^2 = AN^2 + BN^2 = (x_2-x_1)^2 + (y_2-y_1)^2$$

Taking the positive square root, $AB = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$. Hence proved.

Application: $A(7,5)$, $B(2,5)$. $AB = \sqrt{(2-7)^2+(5-5)^2} = \sqrt{25+0} = 5$. Indeed both points have the same ordinate $5$, so they lie on a horizontal line, and the distance equals $|x_2-x_1| = |2-7| = 5$. Verified.

LA 2. Derive the section formula for internal division. A point divides the segment joining $(7,-2)$ and $(1,-5)$ in the ratio $2:1$. Find the point.

Answer: Let $P(x,y)$ divide $A(x_1,y_1)$ and $B(x_2,y_2)$ in the ratio $m:n$ internally. Drop perpendiculars from $A$, $P$, $B$ to the X-axis at $L$, $Q$, $M$. Through $A$ draw a horizontal line meeting $PQ$ extended at $R$ and through $P$ a horizontal line meeting $BM$ at $S$. Then $\triangle ARP \sim \triangle PSB$ (AA similarity). So:

$$\frac{AR}{PS} = \frac{AP}{PB} = \frac{m}{n} \Rightarrow \frac{x-x_1}{x_2 – x} = \frac{m}{n}$$

Cross-multiplying and solving for $x$: $x = \dfrac{m x_2 + n x_1}{m+n}$. Similarly $y = \dfrac{m y_2 + n y_1}{m+n}$.

Application: $A(7,-2)$, $B(1,-5)$, $m:n = 2:1$:

$$P = \left(\tfrac{2(1)+1(7)}{3}, \tfrac{2(-5)+1(-2)}{3}\right) = (3,-4)$$

LA 3. The vertices of a triangle are $A(2,1),B(4,3),C(2,5)$. Verify the property that the centroid divides each median in the ratio $2:1$ from the vertex.

Answer: Centroid $G = \left(\tfrac{2+4+2}{3},\tfrac{1+3+5}{3}\right) = \left(\tfrac{8}{3},3\right)$.

Median from $A$: midpoint of $BC$ is $D = (3,4)$. The point that divides $AD$ in ratio $2:1$ from $A$ is: $\left(\tfrac{2(3)+1(2)}{3},\tfrac{2(4)+1(1)}{3}\right) = \left(\tfrac{8}{3}, 3\right) = G$. Verified.

Median from $B$: midpoint of $AC$ is $E = (2,3)$. Point dividing $BE$ in $2:1$ from $B$: $\left(\tfrac{2(2)+1(4)}{3},\tfrac{2(3)+1(3)}{3}\right) = \left(\tfrac{8}{3}, 3\right) = G$. Verified.

Median from $C$: midpoint of $AB$ is $F = (3,2)$. Point dividing $CF$ in $2:1$ from $C$: $\left(\tfrac{2(3)+1(2)}{3},\tfrac{2(2)+1(5)}{3}\right) = \left(\tfrac{8}{3}, 3\right) = G$. Verified.

All three medians pass through $G$ and $G$ divides each in $2:1$ from the vertex.

LA 4. Show that $A(7,10),B(-2,5),C(3,-4)$ are vertices of an isosceles right triangle. Find the area.

Answer:

$AB^2 = 81+25 = 106$.

$BC^2 = 25+81 = 106$.

$AC^2 = 16+196 = 212$.

Since $AB = BC$ and $AB^2 + BC^2 = 212 = AC^2$, the triangle is right-angled at $B$ and isosceles. Area $= \tfrac{1}{2}\cdot AB\cdot BC = \tfrac{1}{2}\cdot 106 = 53$ sq units.

LA 5. The area of a triangle is $5$ sq units. Two of its vertices are $(2,1)$ and $(3,-2)$. The third vertex lies on the line $y = x+3$. Find the third vertex.

Answer: Let third vertex be $(h, h+3)$. Then:

$\tfrac{1}{2}|2(-2-(h+3))+3((h+3)-1)+h(1+2)| = 5$.

$\Rightarrow |2(-h-5)+3(h+2)+3h| = 10$.

$\Rightarrow |-2h-10+3h+6+3h| = 10 \Rightarrow |4h-4| = 10$.

$4h-4 = \pm 10 \Rightarrow h = \tfrac{7}{2}$ or $h = -\tfrac{3}{2}$.

Third vertices: $\left(\tfrac{7}{2},\tfrac{13}{2}\right)$ or $\left(-\tfrac{3}{2},\tfrac{3}{2}\right)$.

LA 6. Find the coordinates of the points which divide the segment $A(-4,0)$ and $B(0,6)$ into four equal parts.

Answer: Let the dividing points be $P_1, P_2, P_3$ in ratios $1:3$, $1:1$ and $3:1$.

$P_1 = \left(\tfrac{1(0)+3(-4)}{4},\tfrac{1(6)+3(0)}{4}\right) = (-3,1.5)$.

$P_2 = (-2,3)$.

$P_3 = \left(\tfrac{3(0)+1(-4)}{4},\tfrac{3(6)+1(0)}{4}\right) = (-1,4.5)$.

LA 7. Prove that the diagonals of a parallelogram bisect each other using the coordinates $A(0,0),B(a,0),C(a+b,c),D(b,c)$.

Answer: Midpoint of $AC = \left(\tfrac{a+b}{2},\tfrac{c}{2}\right)$. Midpoint of $BD = \left(\tfrac{a+b}{2},\tfrac{c}{2}\right)$. They coincide, so the diagonals bisect each other.

LA 8. The points $A(2,9),B(a,5)$ and $C(5,5)$ are the vertices of a right-angled triangle, right-angled at $B$. Find the value of $a$ and the area of the triangle.

Answer: By Pythagoras $AB^2+BC^2 = AC^2$. $AB^2 = (2-a)^2 + 16$, $BC^2 = (5-a)^2 + 0$, $AC^2 = 9+16 = 25$. So $(2-a)^2 + 16 + (5-a)^2 = 25$. Expanding: $a^2 – 4a + 4 + 16 + a^2 – 10a + 25 = 25 \Rightarrow 2a^2 – 14a + 20 = 0 \Rightarrow a^2 – 7a + 10 = 0 \Rightarrow a = 2 \text{ or } 5$. For $a=2$: $B(2,5)$, $AB = 4$, $BC = 3$, area $= \tfrac{1}{2}\cdot 4\cdot 3 = 6$ sq units. For $a=5$: $B(5,5)$, $AB = \sqrt{9+16}=5$, $BC = 0$ — degenerate, rejected. So $a = 2$ and area $= 6$ sq units.

LA 9. If the points $A(6,1),B(8,2),C(9,4),D(p,3)$ are vertices of a parallelogram taken in order, find $p$.

Answer: Diagonals bisect each other:

$\left(\tfrac{6+9}{2},\tfrac{1+4}{2}\right) = \left(\tfrac{8+p}{2},\tfrac{2+3}{2}\right)$. So $\tfrac{15}{2} = \tfrac{8+p}{2} \Rightarrow p = 7$. Y-coordinates already match.

LA 10. The vertices of a quadrilateral are $A(-3,2),B(5,4),C(7,-6),D(-5,-4)$. Show that $ABCD$ is a parallelogram.

Answer: Midpoint of diagonal $AC = (2,-2)$. Midpoint of diagonal $BD = (0,0)$. They are not equal — so $ABCD$ is NOT a parallelogram. (A useful negative result demonstrating that the question as stated has no solution; the standard textbook variant uses different coordinates that do bisect.)

Re-examining with $A(-3,2),B(5,4),C(7,-6),D(-1,-8)$: midpoint of $AC = (2,-2)$ and midpoint of $BD = (2,-2)$ — diagonals bisect, so it is a parallelogram.

LA 11. Find the third vertex of an equilateral triangle whose two vertices are $(0,0)$ and $(3,\sqrt{3})$.

Answer: Side $= \sqrt{9+3} = 2\sqrt{3}$. Let third vertex be $(x,y)$. Then both distances from $(0,0)$ and $(3,\sqrt 3)$ equal $2\sqrt 3$. So $x^2+y^2 = 12$ and $(x-3)^2+(y-\sqrt 3)^2 = 12$. Subtracting: $6x + 2\sqrt 3 y – 12 = 0 \Rightarrow 3x+\sqrt 3 y = 6 \Rightarrow y = \tfrac{6-3x}{\sqrt 3} = \sqrt 3(2-x)$. Substituting: $x^2 + 3(2-x)^2 = 12 \Rightarrow x^2+12-12x+3x^2 = 12 \Rightarrow 4x^2 – 12x = 0 \Rightarrow x = 0$ or $x=3$. For $x=0$, $y = 2\sqrt 3$; for $x=3$, $y = -\sqrt 3$. Third vertex is $(0, 2\sqrt 3)$ or $(3, -\sqrt 3)$.

Quick-Recall Cheat Sheet

Concept	Formula / Rule
Origin	$O(0,0)$
Distance between $A(x_1,y_1)$ and $B(x_2,y_2)$	$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
Distance from origin	$\sqrt{x^2+y^2}$
Distance from X-axis	$|y|$
Distance from Y-axis	$|x|$
Internal section in ratio $m:n$	$\left(\tfrac{m x_2+n x_1}{m+n},\tfrac{m y_2+n y_1}{m+n}\right)$
Midpoint	$\left(\tfrac{x_1+x_2}{2},\tfrac{y_1+y_2}{2}\right)$
Centroid	$\left(\tfrac{x_1+x_2+x_3}{3},\tfrac{y_1+y_2+y_3}{3}\right)$
Area of triangle	$\tfrac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$
Collinearity test	Area $= 0$
Centroid divides median	$2:1$ from vertex
Diagonals of a parallelogram	Bisect each other (same midpoint)
Square: 4 sides equal + diagonals equal	$ABCD$ is a square
Rhombus: 4 sides equal + diagonals unequal	$ABCD$ is a rhombus
Rectangle: opposite sides equal + diagonals equal	$ABCD$ is a rectangle
Parallelogram: opposite sides equal + diagonals unequal + bisecting	$ABCD$ is a parallelogram

Fill in the Blanks

1. The distance of a point $P(x,y)$ from the origin is _______. Answer: $\sqrt{x^2+y^2}$.
2. The midpoint of the segment joining $(x_1,y_1)$ and $(x_2,y_2)$ is _______. Answer: $\left(\tfrac{x_1+x_2}{2},\tfrac{y_1+y_2}{2}\right)$.
3. Three points are collinear if and only if the area of the triangle formed by them is _______. Answer: zero.
4. The point $(3,0)$ lies on the _______ axis. Answer: X-axis.
5. The centroid divides every median in the ratio _______. Answer: $2:1$.
6. The four regions of the Cartesian plane are called _______. Answer: quadrants.
7. The point of intersection of the X- and Y-axes is the _______. Answer: origin.
8. If $(x,y)$ lies in the third quadrant then both $x$ and $y$ are _______. Answer: negative.

True / False

1. The distance formula gives a positive value always. True.
2. Three points always form a triangle. False — if collinear, they do not.
3. The midpoint formula is a special case of the section formula. True.
4. The origin lies in the first quadrant. False — it is on both axes.
5. The centroid of an equilateral triangle coincides with its incentre and circumcentre. True.
6. The Y-axis is the locus of points whose abscissa is zero. True.

Concept-Check Short Q&A

Q1. What is meant by an ordered pair?

Answer: An ordered pair $(x,y)$ is a pair of numbers in which the order is significant — the first coordinate $x$ is the abscissa and the second coordinate $y$ is the ordinate. The pair $(2,5)$ is different from the pair $(5,2)$.

Q2. What are the coordinates of the origin?

Answer: $(0,0)$.

Q3. What is the locus of points equidistant from two given points?

Answer: The perpendicular bisector of the segment joining the two points.

Q4. Why does the area formula contain a modulus sign?

Answer: Depending on whether the vertices are listed clockwise or anti-clockwise, the algebraic expression may be positive or negative. Since area is always non-negative, we take the modulus.

Q5. What is the distance of any point on the X-axis from the X-axis?

Answer: Zero — points on the X-axis have ordinate $0$.

Q6. State the section formula for external division.

Answer: For external division in ratio $m:n$, the formula is $\left(\dfrac{m x_2 – n x_1}{m-n}, \dfrac{m y_2 – n y_1}{m-n}\right)$.

Q7. If the centroid of a triangle is at the origin, what is the sum of the X-coordinates of its vertices?

Answer: $x_1 + x_2 + x_3 = 0$ (since the X-coordinate of centroid is $\tfrac{x_1+x_2+x_3}{3} = 0$).

Q8. State whether the points $(1,1),(2,2),(3,3)$ form a triangle.

Answer: No — they are collinear (all lie on the line $y=x$), so the area is zero and no triangle is formed.

Q9. What is the maximum area of a triangle that can be inscribed in a circle of radius $r$?

Answer: An equilateral triangle, with area $\tfrac{3\sqrt 3}{4}r^2$. (Beyond Class 10 syllabus but useful trivia.)

Q10. If three points are collinear and we know their X-coordinates and one Y-coordinate, can we always recover the missing two Y-coordinates?

Answer: Not from collinearity alone — collinearity gives one equation; you need either another condition (a slope, another point, an intercept) or the missing values explicitly.

Q11. Find the abscissa of the point of intersection of the X-axis with the segment joining $(2,3)$ and $(2,-5)$.

Answer: The segment is vertical (both X-coordinates are $2$), so it crosses the X-axis at $(2,0)$. Abscissa $= 2$.

Q12. Where does the angle bisector from the origin to the line $y=x$ go?

Answer: The angle bisector of the first quadrant’s right angle at the origin is the line $y=x$ itself — it lies along the diagonal.

Q13. The centroid of a triangle is $(2,3)$ and two of its vertices are $(1,1)$ and $(2,4)$. Find the third.

Answer: $\tfrac{1+2+x_3}{3} = 2 \Rightarrow x_3 = 3$. $\tfrac{1+4+y_3}{3} = 3 \Rightarrow y_3 = 4$. Third vertex $(3,4)$.

Q14. The X-axis is the locus of points whose _______ is zero.

Answer: Y-coordinate (ordinate).

Q15. If a point divides a segment externally in the ratio $1:2$, name a property that distinguishes external from internal division.

Answer: In external division, the dividing point lies outside the segment (on the extension), while in internal division it lies between the two endpoints.

Exam Tips and Strategy

• Sketch first, calculate next. Draw a quick coordinate-plane figure for every problem. It catches sign errors and reveals symmetries.
• Use the appropriate formula. Distance for lengths; section formula for division; area formula for triangles; midpoint as a special case of section.
• Squaring removes ambiguity, but creates extraneous roots. Always check the sign of the answer at the end — distances are positive, but $y$ in equations like $\sqrt{(y+3)^2} = 6$ has two solutions.
• To prove a quadrilateral is a square / rhombus / rectangle / parallelogram, compute all four sides and both diagonals. The combination of equal sides and equal diagonals decides the shape.
• Collinearity? Try the area formula first — it is the fastest test.
• Section formula sign trap. The numerator pairs $m$ with $x_2$ and $n$ with $x_1$ — opposite to the obvious order. Memorise as “outside-times-inside”.
• Centroid shortcut. If asked for the centroid, use the average formula directly — do not compute medians explicitly.
• Equidistant points often translate to the perpendicular bisector — squaring both distances and simplifying gives a clean linear equation in $x$ and $y$.
• Common pitfall: when the question says “the point divides $AB$ such that $AP:PB = 1:2$”, $P$ is closer to $A$. Mixing up $A$ and $B$ flips the ratio.
• Watch the units. If coordinates are in metres, area is in square metres. If they are dimensionless (just numbers), say “square units”.

Frequently Asked Questions (FAQ)

FAQ 1. What is the difference between abscissa and ordinate?

Answer: The abscissa is the X-coordinate (perpendicular distance from the Y-axis); the ordinate is the Y-coordinate (perpendicular distance from the X-axis). For a point $(x,y)$, abscissa $=x$, ordinate $=y$.

FAQ 2. Can the section formula be used for the midpoint?

Answer: Yes — putting $m = n$ (i.e. ratio $1:1$) reduces the section formula to the midpoint formula.

FAQ 3. How do I tell whether a quadrilateral is a square or a rhombus?

Answer: Both have all four sides equal. The square has equal diagonals; the rhombus has unequal diagonals. So compute the two diagonals and compare.

FAQ 4. What does the modulus sign mean in the area formula?

Answer: Modulus (absolute value) gives the non-negative magnitude of the expression inside, since area is always positive regardless of the orientation in which we list the vertices.

FAQ 5. Is the distance formula valid in three dimensions too?

Answer: Yes — for points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ in 3D, the distance is $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$. You will study this in higher classes.

FAQ 6. Why is coordinate geometry important?

Answer: It connects algebra and geometry, allowing geometric problems to be solved using algebraic equations. It underpins computer graphics, physics, GPS navigation, machine learning, and many engineering applications.

FAQ 7. How do I avoid sign errors with negative coordinates?

Answer: Always write subtractions inside parentheses, e.g. $(y_2-y_1)$ becomes $(-3-(-5)) = (-3+5) = 2$, not $-3-5 = -8$.

FAQ 8. Is Exercise 7.4 compulsory for the HSLC examination?

Answer: Exercise 7.4 is marked “optional” in the textbook and is generally not compulsory in the HSLC examination, but the problems strengthen your understanding and similar HOTS-type questions can appear. Going through it is highly recommended.

FAQ 9. What if the area formula gives zero?

Answer: Then the three given points are collinear and do not form a triangle.

FAQ 10. Are collinearity test, slope and area formula equivalent?

Answer: Yes — three points $A$, $B$, $C$ are collinear iff slope of $AB =$ slope of $BC$ iff area of $\triangle ABC = 0$ iff the points satisfy the same line equation. Any of these three tests works.

Glossary of Key Terms

Term	Meaning
Cartesian plane	The plane formed by two perpendicular number lines (X-axis and Y-axis) intersecting at the origin.
Origin	The intersection point of the X- and Y-axes; coordinates $(0,0)$.
Abscissa	The X-coordinate of a point — its perpendicular distance from the Y-axis.
Ordinate	The Y-coordinate of a point — its perpendicular distance from the X-axis.
Quadrants	The four regions into which the axes divide the plane (I, II, III, IV).
Distance Formula	$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ — distance between two points.
Section Formula	Coordinates of a point that divides a line segment in a given ratio $m:n$.
Midpoint	Special case of section formula with ratio $1:1$.
Centroid	Point of intersection of the medians of a triangle; divides each median in ratio $2:1$.
Collinear points	Three or more points lying on the same straight line — area of triangle they form is $0$.
Median	A line segment from a vertex of a triangle to the midpoint of the opposite side.
Area Formula	$\tfrac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$ for a triangle’s area.
Trisection points	Two points that divide a line segment into three equal parts (ratios $1:2$ and $2:1$).
Rhombus	Quadrilateral with all four sides equal but diagonals unequal.
Parallelogram	Quadrilateral with both pairs of opposite sides parallel and equal; diagonals bisect each other.

One-Line Memorisation Aids

• Distance: “Square the differences, add, square-root.”
• Midpoint: “Average the X’s, average the Y’s.”
• Section formula (internal): “Cross the ratios — $m$ goes with $x_2$, $n$ goes with $x_1$, divide by $m+n$.”
• Area: “Each X-coordinate times the difference of the OTHER two Y-coordinates, sum, half-magnitude.”
• Collinearity: “Area zero, points on a line.”
• Centroid: “Average of all three vertex coordinates.”
• Parallelogram test: “Diagonals share the same midpoint.”
• Square vs Rhombus: “All sides equal — diagonals decide.”

Common Mistakes to Avoid

• Forgetting to square the differences before adding in the distance formula.
• Writing the section formula as $\left(\tfrac{m x_1+n x_2}{m+n},\ldots\right)$ — the order is reversed: $m$ with $x_2$.
• Dropping the modulus in the area formula and reporting a negative area.
• Confusing internal and external division — always read the problem carefully.
• Mistaking the X-axis for the Y-axis when finding equidistant points.
• Solving $(y+3)^2 = 36$ as $y+3 = 6$ only — there are two roots, $\pm 6$.
• Reporting the third trisection point as the midpoint — the trisection points are at $1:2$ and $2:1$, neither being the midpoint $1:1$.
• Applying the distance formula across mixed units (e.g. metres and kilometres).
• Forgetting that the centroid divides the median in ratio $2:1$ from the vertex (not from the midpoint).
• Verifying a parallelogram only by equal opposite sides and skipping the diagonal-bisection check.

Chapter Wrap-up

Coordinate Geometry binds together everything you have learnt about points, lines, distances, ratios, and areas — and gives you a single algebraic toolbox to handle them. The three pillars of this chapter are:

1. The Distance Formula — a direct application of Pythagoras’ theorem in the coordinate plane.
2. The Section Formula — derived from similar triangles, with the midpoint as the special case $m=n$.
3. The Area Formula — a clean closed form that doubles as a collinearity test.

If you remember those three formulas and the few derived ideas (centroid, midpoint, parallelogram-diagonal-bisection), you can answer almost every problem the HSLC examination poses. Practise drawing each figure on graph paper, label coordinates clearly, and check your answers by computing distances or midpoints back from your final coordinates — coordinate geometry is one of the few topics in mathematics where every step can be verified geometrically.

The chapter also rewards good algebraic discipline. Many students lose marks not in the formula but in the algebra after the formula — sign errors, mishandled negatives, lost factors of $\tfrac{1}{2}$, mixed-up ratios. Keep your work neat, write the formula above each step, and substitute coordinates inside parentheses.

You have now mastered every formula and every problem of ASSEB Class 10 Mathematics Chapter 7 — Coordinate Geometry. Practice the additional questions, redraw the coordinate-plane figures by hand, and revise the formulas till they feel automatic. For more chapters and exam tips, keep visiting HSLC Guru — wishing you the very best for your HSLC examination!