Question

If a line $L_1$ passing through $A(2,3)$ having inclination $\frac{\pi }{4}$ meets $X-$ axis and $Y-$ axis at $C$ and $D$ respectively and another line $L_2\equiv x+y-7=0$ at point $B$. Then which of the following is/are true

• A. $AB=\sqrt{2}$
• B. $BC=4\sqrt{2}$
• C. mid point of $AD$ is $(1,2)$
• D. $L_1$ is perpendicular to $L_2$

Answer 1

Answer: (A), (B), (C), (D)

$L_1$ is perpendicular to $L_2$

Equation of $L_1$ in parametric form is $\frac{x-2}{\cos \pi /4}=\frac{y-3}{\sin \pi /4}=r$
$\Rightarrow x=2+\frac{r}{\sqrt{2}},y=3+\frac{r}{\sqrt{2}}$

To find $B:2+\frac{r}{\sqrt{2}}+3+\frac{r}{\sqrt{2}}-7=0$
$\Rightarrow r=\sqrt{2}\to AB$
$\therefore B\equiv (3,4)$

Equation of $L_1$ is $y-3=1(x-2)\Rightarrow x-y+1=0$ $\therefore C\equiv (-1,0)$ and $D\equiv (0,1)$
Hence, $AB=\sqrt{2},BC=\sqrt{32}=4\sqrt{2}$

Mid point of $AD=(\frac{2+0}{2},\frac{3+1}{2})=(1,2)$
(Slope of $L_1$ ) (Slope of $L_2)=(+1\left)\right(-1)=-1\Rightarrow L_1\perp L_2$