A ray of light coming from the point P(1, 2) gets reflected from the point Q on the x-axis and then passes through the point R(4, 3). If the point S(h, k) is such that PQRS is a parallelogram, then is equal to: [2024]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
A ray of light coming from the point P(1, 2) gets reflected from the point Q on the x-axis and then passes through the point R(4, 3). If the point S(h, k) is such that PQRS is a parallelogram, then $h k^{2}$ is equal to: [2024]

1 70

2 90

3 60

4 80

Ans.
(1)

Slope of $P' R = \frac{3 - (- 2)}{4 - 1} = \frac{5}{3}$

Equation of $P' R : y + 2 = \frac{5}{3} (x - 1)$

Since, point Q is on x-axis.

Let $Q \equiv (a, 0)$

Also, point Q lies P'R

$\Rightarrow \frac{6}{5} = a - 1 \Rightarrow a = \frac{11}{5}$

Now, PQRS is a parallelogram

$∴ \frac{h + a}{2} = \frac{4 + 1}{2} \Rightarrow h = 5 - \frac{11}{5} = \frac{14}{5}$

and $\frac{2 + 3}{2} = \frac{k}{2} \Rightarrow k = 5$

$∴ h k^{2} = \frac{14}{5} \times 25 = 70$

Want Us to Email you About Special Offers & Updates?