Question

Let $L$ be the line passing through the point $P(1,2)$ such that its intercepted segment between the co-ordinate axes is bisected at $P$. If $L_1$ is line perpendicular to $L$ and passing through the point $(–2,1)$, then the point of intersection of $L$ and $L_1$ is

• A. $(\frac{4}{5},\frac{12}{5})$
• B. $(\frac{3}{5},\frac{23}{10})$
• C. $(\frac{3}{10},\frac{17}{5})$
• D. $(\frac{11}{20},\frac{29}{10})$

Answer 1

The correct option is A $(\frac{4}{5},\frac{12}{5})$


$(1,2)=(\frac{a+0}{2},\frac{b+0}{2})\Rightarrow a=2,b=4$
Equation of the line in intercept form,
$\frac{x}{2}+\frac{y}{4}=1L:2x+y=4$
Line perpendicular to $L$ is $x-2y+k=0$
It passes through $(-2,1)$
$-2-2+k=0\Rightarrow k=4L_1:x-2y+4=0$
Solving $L$ & $L_1,$ we get
$x=\frac{4}{5},y=\frac{12}{5}$
$\therefore$ Point of intersection is $(\frac{4}{5},\frac{12}{5})$