Question

$In$ the figure two lines are intersecting at point (3, 4). The equation of line PA and line PB are respectively

• A. $PA\equiv x-y-1=0andPB\equiv \sqrt{3}x-y+4-\sqrt{3}=0.$
• B. $PA\equiv x-y+1=0andPB\equiv \sqrt{3}x-y-4-\sqrt{3}=0.$
• C. $PA\equiv x-y+1=0andPB\equiv \sqrt{3}x-y+4-\sqrt{3}=0.$
• D. $PA\equiv x+y+1=0andPB\equiv \sqrt{3}x+y+4-\sqrt{3}=0.$

Answer 1

The correct option is C $PA\equiv x-y+1=0andPB\equiv \sqrt{3}x-y+4-\sqrt{3}=0.$

Two lines $\overline{AP}$ and $\overline{BP}$ intersect each other at $(3,4)$.

Now, $\overline{PA}$ makes an angle ${45}^0$ and $\overline{PB}$ makes an angle ${60}^0$ with the X-axis.

We know,

Equation of line: $y-y_1=m(x-x_1)$ where $m=tan\left(\theta \right)$

$\therefore$ Equation of line $\text{PA}$ is,

$y-4=tan\left({45}^0\right)(x-3)$

$y-4=\left(1\right)(X-3)$

$y-4-x+3=0$

$x-y+1=0$

and equation of line $\text{PB}$ is,

$y-4=tan\left({60}^0\right)(x-3)$

$y-4=\sqrt{3}(x-3)$

$y-4-\sqrt{3}x+3\sqrt{3}=0$

$\sqrt{3}x-y+4-3\sqrt{3}=0$

Option $\text{C}$ is correct.