Question

If the image of point $P(2,3)$ in a line $L$ is $Q(4,5)$ then the image of point $R(0,0)$ in the same line is

• A. $(2,2)$
• B. $(4,5)$
• C. $(3,4)$
• D. $(7,7)$

Answer 1

The correct option is D $(7,7)$

We have

$P(x_1,y_1)=(2,3)$

$Q(x_2,y_2)=(4,5)$

then,

mid point of PQ is

$(x_3,y_3)=(\frac{2+4}{2},\frac{3+5}{2})=(3,4)$

now,

slope of $PQ=\frac{y_2-y_1}{x_2-x_1}=\frac{5-3}{4-2}=\frac{2}{2}=1$

slope of line $L=-1$

mid point (3,4) lies on the line L.

then, equation of line L is

$y-4=-1(x-3)$

$\Rightarrow y-4=3-x$

$\Rightarrow x+y-7=\mathrm{0...}\left(1\right)$

Let image point $R(0,0)$ be $S(x_4,y_4)$

Then, mid point of $RS=(\frac{0+x_4}{2},\frac{0+y_4}{2})$

mid point $(\frac{x_4}{2},\frac{y_4}{2})$ lies on the line (1)

So,

$\frac{x_4}{2}+\frac{y_4}{2}=7$

$\Rightarrow x_4+y_4=14...\left(2\right)$

slope of line (2) is

$=\frac{y_4}{x_4}$

Since,

$RS\perp lineL$

So,

$\frac{y_4}{x_4}\times (-1)=-1$

$\Rightarrow x_4=y_4...\left(3\right)$

by equation (2) and (3) to and we get

$x_4=y_4=7$

$(x_4,y_4)=(7,7)$

Hence, the image of R (7,7)

This is the required answer.