SEBA Class 10 Maths Exercise 7.1 Coordinate Geometry

Get free solutions to SEBA Class 10 Maths Exercise 7.1: Coordinate Geometry from SEBA’s new book (2026). We have solved all the problems of the Chapter 7.1 in simple way so that you can understand the concepts and get full marks in your upcoming HSLC matric examination.

If you are a student of SEBA Class 10 and are also looking for reliable solutions for Chapter 7.1 Coordinate Geometry. In this article, you get reliable solutions for the New Book 2026 Class 10 Mathematics Chapter 7.1 Coordinate Geometry in one place.

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SEBA Class 10 Maths Exercise 4.1 Solutions

Question 1

Find the distance between the following pairs of points:

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

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

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

Solution:

(i) Let the given points be $P(2, 3)$ and $Q(4, 1)$.

Then $x_1 = 2, y_1 = 3, x_2 = 4$ and $y_2 = 1$

$\therefore \text{Using the distance formula:} PQ = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$

$= \sqrt{(4 – 2)^2 + (1 – 3)^2}$

$= \sqrt{(2)^2 + (-2)^2} $

$= \sqrt{4 + 4}$

$= \sqrt{8} $

$= \mathbf{2\sqrt{2}\text{ units.}}$

(ii) Let the given points be $P(-5, 7)$ and $Q(-1, 3)$.

Then $x_1 = -5, y_1 = 7, x_2 = -1$ and $y_2 = 3$

$\therefore \text{Distance } PQ = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$

$= \sqrt{(-1 + 5)^2 + (3 – 7)^2}$

$= \sqrt{(4)^2 + (-4)^2} $

$= \sqrt{16 + 16}$

$= \sqrt{32} $

$= \mathbf{4\sqrt{2}\text{ units.}}$

(iii) Let the given points be $P(a, b)$ and $Q(-a, -b)$.

Then $x_1 = a, y_1 = b, x_2 = -a$ and $y_2 = -b$

$\therefore \text{Distance } PQ = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$

$= \sqrt{(-a – a)^2 + (-b – b)^2}$

$= \sqrt{(-2a)^2 + (-2b)^2} $

$= \sqrt{4a^2 + 4b^2}$

$= \sqrt{4(a^2 + b^2)} $

$= \mathbf{2\sqrt{a^2 + b^2}\text{ units.}}$

2. 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?

Solution:

Let the points be $A(0, 0)$ and $B(36, 15)$.

Using the distance formula: $AB = \sqrt{(36 – 0)^2 + (15 – 0)^2}$

$= \sqrt{36^2 + 15^2}$

$= \sqrt{1296 + 225}$

$= \sqrt{1521} = 39$

Yes, taking town A at the origin $(0, 0)$ and town B at $(36, 15)$, the distance between town A and town B is $39\text{ km}$.

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SEBA Class 10 Advanced Maths Exercise 3.3 Solutions

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

Solution:

Let the points be $A(1, 5)$, $B(2, 3)$, and $C(-2, -11)$.

Using the distance formula:

$AB = \sqrt{(2 – 1)^2 + (3 – 5)^2} $

$= \sqrt{1^2 + (-2)^2} $

$= \sqrt{1 + 4} $

$= \sqrt{5}$

$BC = \sqrt{(-2 – 2)^2 + (-11 – 3)^2} $

$= \sqrt{(-4)^2 + (-14)^2} $

$= \sqrt{16 + 196} $

$= \sqrt{212} $

$= 2\sqrt{53}$

$AC = \sqrt{(-2 – 1)^2 + (-11 – 5)^2} $
$= \sqrt{(-3)^2 + (-16)^2} $
$= \sqrt{9 + 256} $
$= \sqrt{265}$

Since $AB + BC \ne AC$, the points A, B, and C are not collinear.

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

Solution:

Let the points be $A(5, -2)$, $B(6, 4)$, and $C(7, -2)$.

Using the distance formula, we have;

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

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

$AC = \sqrt{(7 – 5)^2 + [-2 – (-2)]^2} $
$= \sqrt{2^2 + 0^2} $
$= \sqrt{4} = 2$

Since $AB = BC = \sqrt{37} \ne AC$, triangle ABC is an isosceles triangle.

5. In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 7.9. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.

Solution:

From Fig. 7.9, the coordinates of the points are $A(3, 4)$, $B(6, 7)$, $C(9, 4)$, and $D(6, 1)$.

Then by distance formula, we have

$AB = \sqrt{(6 – 3)^2 + (7 – 4)^2} $
$= \sqrt{3^2 + 3^2} $
$= \sqrt{9 + 9} $
$= \sqrt{18} $
$= 3\sqrt{2}$

$BC = \sqrt{(9 – 6)^2 + (4 – 7)^2} $
$= \sqrt{3^2 + (-3)^2} $
$= \sqrt{9 + 9} $
$= \sqrt{18} $
$= 3\sqrt{2}$

$CD = \sqrt{(6 – 9)^2 + (1 – 4)^2} $
$= \sqrt{(-3)^2 + (-3)^2} $
$= \sqrt{9 + 9} $
$= \sqrt{18} $
$= 3\sqrt{2}$

$DA = \sqrt{(3 – 6)^2 + (4 – 1)^2} $
$= \sqrt{(-3)^2 + 3^2} $
$= \sqrt{9 + 9} $
$= \sqrt{18} $
$= 3\sqrt{2}$

Also, diagonals
$AC = \sqrt{(9 – 3)^2 + (4 – 4)^2} $
$= \sqrt{6^2 + 0^2} = 6$

$BD = \sqrt{(6 – 6)^2 + (1 – 7)^2} $
$= \sqrt{0^2 + (-6)^2} = 6$

Since all four sides are equal ($AB = BC = CD = DA = 3\sqrt{2}$) and both diagonals are equal ($AC = BD = 6$).

Thus, ABCD is a square. Therefore, Champa is correct.

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

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

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

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

Solution:

(i) Let the given points be $A(-1, -2)$, $B(1, 0)$, $C(-1, 2)$, and $D(-3, 0)$.

Using the distance formula, we have;

$AB = \sqrt{[1 – (-1)]^2 + [0 – (-2)]^2} $
$= \sqrt{2^2 + 2^2} $
$= \sqrt{8} $
$= 2\sqrt{2}$

$BC = \sqrt{(-1 – 1)^2 + (2 – 0)^2} $
$= \sqrt{(-2)^2 + 2^2} $
$= \sqrt{8} $
$= 2\sqrt{2}$

$CD = \sqrt{[-3 – (-1)]^2 + (0 – 2)^2} $
$= \sqrt{(-2)^2 + (-2)^2} $
$= \sqrt{8} $
$= 2\sqrt{2}$

$DA = \sqrt{[-1 – (-3)]^2 + [-2 – 0]^2} $
$= \sqrt{2^2 + (-2)^2} $
$= \sqrt{8} $
$= 2\sqrt{2}$

Again,

$AC = \sqrt{[-1 – (-1)]^2 + [2 – (-2)]^2} $
$= \sqrt{0^2 + 4^2} $
$= 4$

$BD = \sqrt{(-3 – 1)^2 + (0 – 0)^2} $
$= \sqrt{(-4)^2 + 0} $
$= 4$

Since all four sides are equal, i.e., $AB = BC = CD = DA = 2\sqrt{2}$, and

both diagonals are equal ($AC = BD = 4$).

Hence, the quadrilateral ABCD is a Square.

(ii) Let the given points be $A(-3, 5)$, $B(3, 1)$, $C(0, 3)$, and $D(-1, -4)$.

Using the distance formula, we have;

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

$= \sqrt{6^2 + (-4)^2}$

$= \sqrt{36 + 16}$

$= \sqrt{52}$

$= 2\sqrt{13}$

$BC = \sqrt{(0 – 3)^2 + (3 – 1)^2}$

$= \sqrt{(-3)^2 + 2^2}$

$= \sqrt{9 + 4}$

$= \sqrt{13}$

$CD = \sqrt{(-1 – 0)^2 + (-4 – 3)^2}$

$= \sqrt{(-1)^2 + (-7)^2}$

$= \sqrt{1 + 49}$

$= \sqrt{50}$

$= 5\sqrt{2}$

$DA = \sqrt{[-3 – (-1)]^2 + [5 – (-4)]^2}$

$= \sqrt{(-2)^2 + 9^2}$

$= \sqrt{4 + 81}$

$= \sqrt{85}$

Again,

$AC = \sqrt{[0 – (-3)]^2 + (3 – 5)^2}$

$= \sqrt{3^2 + (-2)^2}$

$= \sqrt{9 + 4}$

$= \sqrt{13}$

Here, $AC + BC = \sqrt{13} + \sqrt{13} = 2\sqrt{13} = AB$.

Therefore the points $A$, $B$, and $C$ are collinear.

Therefore, the given points cannot form a general quadrilateral.

(iii) Let the given points be $A(4, 5)$, $B(7, 6)$, $C(4, 3)$, and $D(1, 2)$.

Using the distance formula, we have;

$AB = \sqrt{(7 – 4)^2 + (6 – 5)^2}$

$= \sqrt{3^2 + 1^2}$

$= \sqrt{9 + 1}$

$= \sqrt{10}$

$BC = \sqrt{(4 – 7)^2 + (3 – 6)^2}$

$= \sqrt{(-3)^2 + (-3)^2}$

$= \sqrt{9 + 9}$

$= \sqrt{18}$

$= 3\sqrt{2}$

$CD = \sqrt{(1 – 4)^2 + (2 – 3)^2}$

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

$= \sqrt{9 + 1}$

$= \sqrt{10}$

$DA = \sqrt{(4 – 1)^2 + (5 – 2)^2}$

$= \sqrt{3^2 + 3^2}$

$= \sqrt{9 + 9}$

$= \sqrt{18}$

$= 3\sqrt{2}$

Again,

$AC = \sqrt{(4 – 4)^2 + (3 – 5)^2}$

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

$= \sqrt{4}$

$= 2$

$BD = \sqrt{(1 – 7)^2 + (2 – 6)^2}$

$= \sqrt{(-6)^2 + (-4)^2}$

$= \sqrt{36 + 16}$

$= \sqrt{52}$

$= 2\sqrt{13}$

Since opposite sides are equal, i.e., $AB = CD = \sqrt{10}$ and $BC = DA = 3\sqrt{2}$, but the diagonals are not equal ($AC \ne BD$).

Hence, the quadrilateral ABCD is a Parallelogram.

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

Solution:

Let $A(2, -5)$ and $B(-2, 9)$ be the given points.

Also, let $P(x, 0)$ be the point on the $x$-axis.

Since $P$ is equidistant from $A$ and $B$,

$\therefore AP = BP $

$\implies AP^2 = BP^2$.

$\implies ({\sqrt{(x – 2)^2 + [0 – (-5)]^2}})^2 = ({\sqrt{[x – (-2)]^2 + (0 – 9)^2}})^2 $

$\implies (x – 2)^2 + [0 – (-5)]^2 = [x – (-2)]^2 + (0 – 9)^2$

$\implies x^2 – 4x + 4 + 25 = x^2 + 4x + 4 + 81$

$\implies -4x + 29 = 4x + 85$

$\implies -8x = 56$

$\implies x = -7$

Therefore, the required point is $(-7, 0)$.

Question 8

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

Solution:

$P(2, -3)$ and $Q(10, y)$ are the given points.

Also, the distance between the points $PQ = 10\text{ units}$.

$\therefore PQ = 10$

$\implies PQ^2 = 10^2$

$\implies ({\sqrt{(10 – 2)^2 + [y – (-3)]^2}})^2 = 100$

$\implies (10 – 2)^2 + [y – (-3)]^2 = 100$

$\implies (8)^2 + (y + 3)^2 = 100$

$\implies 64 + y^2 + 6y + 9 = 100$

$\implies y^2 + 6y + 73 = 100$

$\implies y(y + 9) – 3(y + 9) = 0$

$\implies (y + 9)(y – 3) = 0$

$\implies y + 9 = 0 \quad \text{or} \quad y – 3 = 0$

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

Therefore, the required values of $y$ are $3$ and $-9$.

Question 9

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.

Solution:

Let $P(5, -3)$ and $R(x, 6)$ be the given points.
Also, let $Q(0, 1)$ be the point equidistant from $P$ and $R$.
Since $Q$ is equidistant from $P$ and $R$,

$\therefore QP = QR$
$\implies QP^2 = QR^2$
$\implies ({\sqrt{(5 – 0)^2 + (-3 – 1)^2}})^2 = ({\sqrt{(x – 0)^2 + (6 – 1)^2}})^2$
$\implies (5 – 0)^2 + (-3 – 1)^2 = (x – 0)^2 + (6 – 1)^2$
$\implies 5^2 + (-4)^2 = x^2 + 5^2$
$\implies 25 + 16 = x^2 + 25$
$\implies x^2 = 16$
$\implies x = \pm 4$

Thus, the values of $x$ are either $4$ and $-4$.

When $x = 4$

$QR = \sqrt{(x – 0)^2 + (6 – 1)^2}$

$= \sqrt{( 4 – 0)^2 + 5^2}$

$= \sqrt{16 + 25}$

$= \sqrt{41}$

And,

$PR = \sqrt{(4 – 5)^2 + [6 – (-3)]^2}$

$= \sqrt{(-1)^2 + 9^2}$

$= \sqrt{1 + 81}$

$= \sqrt{82}$

When $x = -4$:

$QR = \sqrt{(x – 0)^2 + (6 – 1)^2}$

$= \sqrt{(- 4 – 0)^2 + 5^2}$

$= \sqrt{16 + 25}$

$= \sqrt{41}$

And,

$PR = \sqrt{(-4 – 5)^2 + [6 – (-3)]^2}$

$= \sqrt{(-9)^2 + 9^2}$

$= \sqrt{81 + 81}$

$= \sqrt{162}$

$= 9\sqrt{2}$

Therefore, the values of $x$ are $\pm 4$, $QR = \sqrt{41}$, and $PR = \sqrt{82} \text{ or } 9\sqrt{2}$.

Question 10

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

Solution:

Let $A(3, 6)$ and $B(-3, 4)$ be the given points.

Also, let $P(x, y)$ be the point equidistant from $A$ and $B$.

Since $P$ is equidistant from $A$ and $B$,

$\therefore AP = BP$

$\implies AP^2 = BP^2$

$\implies ({\sqrt{(x – 3)^2 + (y – 6)^2}})^2 = ({\sqrt{[x – (-3)]^2 + (y – 4)^2}})^2$

$\implies (x – 3)^2 + (y – 6)^2 = [x – (-3)]^2 + (y – 4)^2$

$\implies x^2 – 6x + 9 + y^2 – 12y + 36 = x^2 + 6x + 9 + y^2 – 8y + 16$

$\implies -6x – 12y + 45 = 6x – 8y + 25$

$\implies -6x – 6x – 12y + 8y + 45 – 25 = 0$

$\implies -12x – 4y + 20 = 0$

$\implies -4(3x + y – 5) = 0$

$\implies 3x + y – 5 = 0$

Therefore, the required relation between $x$ and $y$ is $3x + y – 5 = 0$.

Question 11

The coordinates of a point which lies on the y-axis and 5 units away from the origin can be:

(i) $(0, 5)$
(ii) $(5, 0)$
(iii) $(0, -5)$
(iv) $(-5, 0)$

Choose the correct alternatives –
(a) (i) and (iii)
(b) (i) and (iv)
(c) (ii) and (iv)
(d) (iii) and (iv)

Solution:

A point on the $y$-axis has an $x$-coordinate of $0$ (i.e., in the form $(0, y)$). At a distance of $5$ units from the origin, $y = \pm 5$. Thus, the coordinates are $(0, 5)$ or $(0, -5)$, which corresponds to (i) and (iii).

Correct Options: (a) (i) and (iii)

Question 12

If the three vertices of a triangle are $A(1, 1)$, $B(4, 1)$ and $C(4, 4)$, then the triangle formed by them is:

(i) right angled triangle
(ii) equilateral triangle
(iii) Isosceles triangle
(iv) scalene triangle

Choose the correct option –
(a) (ii) and (iv)
(b) (iii) and (iv)
(c) (i) and (iii)
(d) (i) and (iv)

Solution:

$AB = \sqrt{(4 – 1)^2 + (1 – 1)^2} = \sqrt{3^2 + 0} = 3$

$BC = \sqrt{(4 – 4)^2 + (4 – 1)^2} = \sqrt{0 + 3^2} = 3$

$AC = \sqrt{(4 – 1)^2 + (4 – 1)^2} = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}$

Since $AB = BC = 3$, it is an isosceles triangle.

Also, $AB^2 + BC^2 = 9 + 9 = 18 = AC^2$, which means it is a right-angled triangle.

Correct Option: (c) (i) and (iii)

Question 13

In the following question a statement of assertion (A) is followed by a statement of reason (R)

Assertion (A): Point $P(0, 2)$ is the point of intersection of y-axis with the line $3x + 2y = 4$.

Reason (R): The distance of the point $P(0, 2)$ from x-axis is 2 units.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (А).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true

Solution:

• Assertion (A): For intersection with the $y$-axis, substitute $x = 0$ into $3x + 2y = 4$:

  $3(0) + 2y = 4 $
  $\implies 2y = 4 $
  $\implies y = 2$

  Thus, $P(0, 2)$ is the point of intersection. (Assertion is True)

• Reason (R): The distance of any point $(x, y)$ from the $x$-axis is $\vert{}y\vert{} = \vert{}2\vert{} = 2\text{ units}$. (Reason is True)

• However, Reason (R) does not explain why $P(0, 2)$ is the point of intersection of the given line.

• Correct Option: (b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

Question 14

Match the lines with their equations:

(a) A→P, В →Q, С→R
(b) A→R, B → P, C →Q
(c) A→R, B→Q, C→Р
(d) A→Q, B → R, C P

Solution:

• (A) Line parallel to $x$-axis has an equation of the form $y = k \implies y = 4$ (Q)

• (B) Line parallel to $y$-axis has an equation of the form $x = k \implies x = -2$ (R)

• (C) Line passing through the origin has no constant term $\implies y = mx$ (P)

• Correct Alternative: (d) $A \rightarrow Q, B \rightarrow R, C \rightarrow P$

Question 15

Statement (i): The perimeter of a triangle is $2 + \sqrt{2}$ whose coordinates of the vertices are $(1, 0)$, $(0, 1)$ and $(-1, 0)$.

Statement (ii): The coordinates of the mid-point of a line segment joining the points $A(x_1, y_1)$ and $B(x_2, y_2)$ is $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$.

(a) Both Statement (i) and Statement (ii) are true
(b) Both Statement (i) and Statement (ii) are false
(c) Statement (i) is true but Statement (ii) is false
(d) Statement (i) is false but Statement (ii) is true

Solution:

• Statement (i):
  Let $A(1, 0)$, $B(0, 1)$, and $C(-1, 0)$.
  $AB = \sqrt{(0 – 1)^2 + (1 – 0)^2} = \sqrt{1 + 1} = \sqrt{2}$
  $BC = \sqrt{(-1 – 0)^2 + (0 – 1)^2} = \sqrt{1 + 1} = \sqrt{2}$
  $CA = \sqrt{(1 – (-1))^2 + (0 – 0)^2} = \sqrt{2^2} = 2$
  $\text{Perimeter} = AB + BC + CA $
  $= \sqrt{2} + \sqrt{2} + 2 $
  $= 2 + 2\sqrt{2} \ne 2 + \sqrt{2}$
  Hence, Statement (i) is False.

• Statement (ii): The standard midpoint formula is $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$, which is True.

• Correct Option: (d) Statement (i) is false but Statement (ii) is true

Question 16

The distance between the points $A(x, -1)$ and $B(3, -2)$ is $x + 5$, then the value of x is.

Arrange the randomly given steps in proper order:

(i) $9 + 1 – 25 = 16x$

(ii) $x = -\frac{15}{16}$

(iii) $\sqrt{(3 – x)^2 + \{(-2) – (-1)\}^2} = x + 5$

(iv) $9 – 6x + x^2 + (-1)^2 = x^2 + 10x + 25$

(a) (iii)→(i)→(iv)→(ii)
(b) (iii)→(iv)→(i)→(ii)
(c) (iv)→(iii)→(i)→(ii)
(d) (iv)→(iii)→(ii)→(i)

Solution:

1. Step 1: Applying the distance formula:

   $\sqrt{(3 – x)^2 + \{(-2) – (-1)\}^2} = x + 5 \quad \rightarrow \textbf{(iii)}$

2. Step 2: Squaring both sides and expanding:

   $9 – 6x + x^2 + (-1)^2 $
   $= x^2 + 10x + 25 \quad \rightarrow \textbf{(iv)}$

3. Step 3: Rearranging like terms:

   $9 + 1 – 25 = 16x \quad \rightarrow \textbf{(i)}$

4. Step 4: Solving for $x$:

   $x = -\frac{15}{16} \quad \rightarrow \textbf{(ii)}$

• Correct Option: (b) (iii) $\rightarrow$ (iv) $\rightarrow$ (i) $\rightarrow$ (ii)