Find the image of the point (-1, 5, 2) in the line (2x-4)/(2) = y/2 = (2-z)/(3). Find the length of the line segment joining the points (given point and the image point).

MathsCBSE Class 12 · Board exam 2025, set 65/7/1 · 5 marks · Long answer

CBSE Class 12 Mathematics 2025: question 35(b)

Set in the CBSE Class 12 board exam of 2025, paper 65/7/1, section D, question 35(b), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q35(b) · 5 marks · Long answerThree Dimensional Geometry

Find the image of the point $(- 1, 5, 2)$ in the line $\frac{2 x - 4}{2} = \frac{y}{2} = \frac{2 - z}{3}$ . Find the length of the line segment joining the points (given point and the image point).

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 65/7/1.

Hints

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

Hint 1

Rewrite $\frac{2 - z}{3}$ as $\frac{z - 2}{- 3}$ and simplify $\frac{2 x - 4}{2}$ to get the standard form.

Hint 2

Find the foot of the perpendicular M by making AM perpendicular to the line's direction.

Hint 3

The foot M is the mid-point of the point and its image.

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.

Standard form of the line: $\frac{x - 2}{1} = \frac{y}{2} = \frac{z - 2}{- 3}$ , with direction ratios $1, 2, - 3$ . Let $A (- 1, 5, 2)$ , let $A^{'} (a, b, c)$ be its image and let $M$ be the foot of the perpendicular from $A$ to the line. Then $M$ is the mid-point of $A A^{'}$ .
1 mark
Any point on the line is $M (λ + 2, 2 λ, - 3 λ + 2)$ .
½ mark
Direction ratios of $A M$ : $λ + 3, 2 λ - 5, - 3 λ$ .
½ mark
Since $A M ⊥$ line: $1 (λ + 3) + 2 (2 λ - 5) - 3 (- 3 λ) = 0 \Rightarrow 14 λ - 7 = 0 \Rightarrow λ = \frac{1}{2}$ . So $M (\frac{5}{2}, 1, \frac{1}{2})$ .
1 mark
$M$ is the mid-point of $A A^{'}$ : $\frac{a - 1}{2} = \frac{5}{2}, \frac{b + 5}{2} = 1, \frac{c + 2}{2} = \frac{1}{2} \Rightarrow a = 6, b = - 3, c = - 1$ .
1 mark
Image of the point: $A^{'} (6, - 3, - 1)$ .
½ mark
Length $A A^{'} = (6 + 1)^{2} + (- 3 - 5)^{2} + (- 1 - 2)^{2} = 49 + 64 + 9 = 122$ units.
½ mark

Answer: Image $(6, - 3, - 1)$ ; length of segment $= 122$ units

Where the marks are

Writes the line in standard form $\frac{x - 2}{1} = \frac{y}{2} = \frac{z - 2}{- 3}$ (1 mark)
General point $M (λ + 2, 2 λ, - 3 λ + 2)$ (½ mark)
Direction ratios of AM (½ mark)
Uses perpendicularity to get $λ = \frac{1}{2}$ (1 mark)
Uses mid-point condition to get $a = 6, b = - 3, c = - 1$ (1 mark)
States image $(6, - 3, - 1)$ (½ mark)
Finds $A A^{'} = 122$ (½ mark)

Revise the chapter

Chapter: Three Dimensional Geometry (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Three Dimensional Geometry, one at a time, marked against the answer key and marking points.Back to the full paper2025 board paper, set 65/7/1: every question, by section.

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CBSE Class 12 Mathematics 2025: question 35(b)

Hints

Step-by-step solution

Revise the chapter

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