MathsCBSE Class 10 · Basic · Board exam 2025, set 430/6/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Basic 2025: question 38

Set in the CBSE Class 10 board exam of 2025, paper 430/6/1, section E, question 38, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q38 · 4 marks · Case-basedCompetency-basedCoordinate Geometry

Gurveer and Arushi built a robot that can paint a path as it moves on a graph paper. Some co-ordinates of points are marked on it. It starts from (0, 0), moves to the points listed in order (in straight lines) and ends at (0, 0).

Arushi entered the points P(8, 6), Q(12, 2) and S( $- 6$ , 6) in order. The path drawn by robot is shown in the figure.

Based on the above, answer the following questions:

(i)
Determine the distance OP.
1 mark
(ii)
QS is represented by equation $2 x + 9 y = 42$ . Find the co-ordinates of the point where it intersects y-axis.
1 mark
(iii)(a)
Point R(4.8, y) divides the line segment OP in a certain ratio, find the ratio. Hence, find the value of y.
2 marks
OR
(iii)(b)
Using distance formula, show that $\frac{P Q}{O S} = \frac{2}{3}$ .
2 marks

Figure in the paper: Coordinate graph with x-axis and y-axis through origin O. Points S(-6, 6) in the second quadrant, P(8, 6) in the first quadrant, and Q(12, 2) in the first quadrant. The robot path goes O to S (arrow), S to Q (straight line, arrow pointing left toward S), Q to P (arrow), and P to O (through point R on the line, arrow), with R marked on segment OP. A small robot illustration is shown at left. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 430/6/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Use the distance formula $(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ ; on the y-axis, x = 0.

Hint 2

For R, take the ratio as k : 1 and use only the x-coordinate 4.8 to find k, then find y.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) $O P = (8 - 0)^{2} + (6 - 0)^{2} = 64 + 36 = 100 = 10$ units.
1 mark
(ii) On the y-axis $x = 0$ : $9 y = 42 \Rightarrow y = \frac{14}{3}$ . Point of intersection is $(0, \frac{14}{3})$ .
1 mark
(iii)(a) Let R divide OP in the ratio $k : 1$ . By the section formula, $x$ -coordinate: $4.8 = \frac{8 k + 0}{k + 1} \Rightarrow 4.8 k + 4.8 = 8 k \Rightarrow 3.2 k = 4.8 \Rightarrow k = \frac{3}{2}$ . So $O R : R P = 3 : 2$ .
1½ marks
(iii)(a) $y = \frac{3 \times 6 + 2 \times 0}{3 + 2} = \frac{18}{5} = 3.6$ .
½ mark
(iii)(b) $P Q = (12 - 8)^{2} + (2 - 6)^{2} = 16 + 16 = 32 = 42$ .
½ mark
(iii)(b) $O S = (- 6 - 0)^{2} + (6 - 0)^{2} = 36 + 36 = 72 = 62$ .
½ mark
(iii)(b) $\frac{P Q}{O S} = \frac{4 2}{6 2} = \frac{2}{3}$ . Hence shown.
1 mark

Answer: (i) 10 units; (ii) $(0, \frac{14}{3})$ ; (iii)(a) 3 : 2, $y = \frac{18}{5}$ ; (iii)(b) $P Q = 42$ , $O S = 62$ , ratio $\frac{2}{3}$

Where the marks are

(i) OP = 10 using distance formula (1 mark)
(ii) Intersection with y-axis $(0, \frac{14}{3})$ (1 mark)
(iii)(a) Section formula on x-coordinate giving ratio 3 : 2 (1½ marks)
(iii)(a) $y = \frac{18}{5}$ (½ mark)
(iii)(b) $P Q = 32$ (½ mark), $O S = 72$ (½ mark), ratio $\frac{2}{3}$ (1 mark)

Revise the chapter

Chapter: Coordinate Geometry (notes, formulas, where marks are lost and every past question)
Topic: Applications of the Distance Formula

Practise this chapterBoard-style questions from Coordinate Geometry, one at a time, marked against the answer key and marking points.Back to the full paper2025 board paper, set 430/6/1 (Basic): every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 10 Mathematics Basic 2025: question 38

Hints

Step-by-step solution

Revise the chapter

More questions from Coordinate Geometry