Question

The equation of the sides $AB,BC,CA$ of a $△ABC$ are $2x-y-3=0,6x+y-21=0$ and $2x+y-5=0$ respectively. The external bisector of angle $A$ passes through the point

• A. $(3,1)$
• B. $(4,2)$
• C. $(2,-7)$
• D. $(2,1)$

Answer 1

Answer: (D)

$(2,1)$

The equation of sides $AB,BC,CA$ of $△ABC$ are,

$2x-y-3=0$ ---(i)

$6x+y-21=0$ ---(ii)

$2x+y-5=0$ ---(iii)

Now from $\left(i\right)$ and $\left(iii\right)$ we get

$4-1=3>0$ i.e.$[a_1{a}_2+b_1{b}_2]$

As $a_1{a}_2+b_1{b}_2>0$

We conclude bisector of A is,

$\frac{2x-y-3}{\sqrt{5}}=-\frac{2x+y-5}{\sqrt{5}}$

$2x-y-3=-2x-y+5$

$x=2$

For $x=2$ we have $y=1$ putting value in $\left(i\right)$

Therefore the external angle bisector of $A$ passes through $(2,1)$

Option $\text{D}$ is the correct answer