Mathematics · Straight lines

JEE Main 2024 — 8 April, Shift 2 — Question 25

Let a ray of light passing through the point $(3,10)$ reflects on the line $2 x+y=6$ and the reflected ray passes through the point $(7,2)$ . If the equation of the incident ray is $a x+b y+1=0$ , then $\mathrm{a}^{2}+\mathrm{b}^{2}+3 \mathrm{ab}$ is equal to

Answer: 1

Numerical answer — enter this value.

Step-by-step solution

For $\mathrm{B}^{\prime}, \quad \frac{\mathrm{x}-7}{2}=\frac{\mathrm{y}-2}{1}=-2\left(\frac{14+2-6}{5}\right)$

$\frac{x-7}{2}=\frac{y-2}{1}=-4$

$x=-1, \quad y=-2, \quad B^{\prime}(-1,-2)$

incident ray ${AB}$

$\mathrm{M}_{\mathrm{AB}^{\prime}}=3$

$y+2=3(x+1)$

$3 \mathrm{x}-\mathrm{y}+1=0$

$\mathrm{a}=3 \mathrm{~b}=-1$

$a^{2}+b^{2}+3 a b=9+1-9=1$

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Exam: JEE Main 2024
Paper: 8 April, Shift 2
Subject: Mathematics
Chapter: Straight lines
Topic: Angle between lines, perpendicular distance & distance between parallel lines, foot, image

If the line segment joining the points (5,2) and (2, a) subtends an angle pi/4 at the origin, then the absolute value of the product of all…

Answer: 1

Step-by-step solution

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