JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Straight Lines

A ray of light passing through the point

P (2, 3)

reflects on the x-axis at point A and the reflected ray passes through the point

Q (5, 4)

. Let R be the point that divides the line segment AQ internally into the ratio

2 : 1

. Let the co-ordinates of the foot of the perpendicular M from R on the bisector of the angle PAQ be

(α, β)

. Then, the value of

7 α + 3 β

is equal to ____.

Answer: 31

Solution

To find the reflection point A on the x-axis, we use the image of P in the x-axis, which is

P^{'} (2, - 3)

. The reflected ray from A passes through Q, so A lies on the line

P^{'} Q

.

The equation of line

P^{'} Q

is:

\frac{y + 3}{x - 2} = \frac{4 + 3}{5 - 2} = \frac{7}{3}

This gives

3 y + 9 = 7 x - 14

, or

7 x - 3 y = 23

.

Setting

y = 0

(since A is on the x-axis):

x = \frac{23}{7}

, so

A (\frac{23}{7}, 0)

and

α = \frac{23}{7}

.

Point R divides AQ in ratio

2 : 1

, so the y-coordinate of R is:

β = \frac{2 ( 4 ) + 1 ( 0 )}{2 + 1} = \frac{8}{3}

Therefore:

7 α + 3 β = 7 (\frac{23}{7}) + 3 (\frac{8}{3}) = 23 + 8 = 31

JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Integer answer