Looking for accurate SEBA Class 10 Maths Exercise 7.1 (Coordinate Geometry) textual solutions for the 2026–27 academic year? Access complete step-by-step solutions covering the distance formula d = sqrt((x2 – x1)^2 + (y2 – y1)^2), collinearity tests, finding points equidistant from given coordinates, and geometric shape verification to score full marks in your HSLC board exams.
EXERCISE 7.1
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)
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.
SEBA Class 10 Maths Exercise 7.1 Solutions
3. Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.
4. Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.
Class 10 Maths Chapter 7 Exercise 7.1 Question Answer SEBA 2026-27
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.
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)
7. Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).
Distance Formula Class 10 SEBA Ex 7.1 Solutions
8. Find the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.
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.
10. Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (-3, 4).
SEBA Class 10 Maths Chapter 7 Exercise 7.1 Textbook Answers
11. The cordinates 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)
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) scalen triangle
Choose the correct option –
(a) (ii) and (iv) (b) (iii) and (iv)
(c) (i) and (iii) (d) (i) and (iv)
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.
Choose the correct option from the given alternatives.
(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 (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true
Class 10 Coordinate Geometry Exercise 7.1 HSLC Solutions
14. Match the lines with their equatioins
Column I (line type)
Column II (equation example)
(A) Line parallel to x-axis
(P) y = mx
(B) Line parallel to y-axis
(Q) y = 4
(C) Line passing through the origin
(R) x = -2
Choose the correct alternatives:
(a) A -> P, B -> Q, C -> R (b) A -> R, B -> P, C -> Q
(c) A -> R, B -> Q, C -> P (d) A -> Q, B -> R, C -> P
15. Two statements are given below:
Statement (i) : The perimeter of a triangle is 2 + √2 whose co-ordinates of the vertices are (1, 0), (0, 1) and (-1, 0)
Statement (ii) : The co-ordinates of the mid-point of a line segment joining the points A (x1, y1) and B (x2, y2) is ((x1 + x2)/2, (y1 + y2)/2)
Choose the correct alternatives:
(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
16. The distance between the points A (x, -1) and B (3, -2) is x+5, then the value of x is.
The steps for finding the value of x are given below randomly. Arrange the steps in proper order.
(i) 9 + 1 – 25 = 16x
(ii) x = -15/16
(iii) √((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)
📐 Updated Solutions Notice: This page features complete, step-by-step SEBA Class 10 Mathematics Chapter 7 (Coordinate Geometry) Exercise 7.1 Textual Solutions updated for the 2026–27 academic session. All distance formula evaluations, collinearity checks, and vertex proofs for triangles and quadrilaterals strictly follow the latest revised Assam Board (SEBA) HSLC curriculum.
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