Question

in triange ABC the cordinates of A are (4 , -1) line x-y-1=0 and 2x-y = 3 are internal bisectors of angle B and C. Then radius of the in circle of triangle ABC is

Answer 1

Dear Student,

Let the point of intersection of angle bisectors of angle B and C be I i.e. the incentre of triangle ABC

Now, solving the two equations to get point P
$\Rightarrow$x - y = 1 .................. (1)
$\Rightarrow$2x - y = 3 .................. (2)
Subtracting (1) from (2) $\Rightarrow$
$\Rightarrow$2x - y - (x - y) = 3 - 1
$\Rightarrow$2x - y - x + y = 2
$\Rightarrow$2x - x = 2
$\Rightarrow$x = 2
From (1) $\Rightarrow$
$\Rightarrow$2 - y = 1
$\Rightarrow$y = 1
Therefore, co-ordinates of I are (2 , 1)

Now radius of circle is equal to the distance between A and I
$\Rightarrow$radius = AI = $\sqrt{{\left(4-2\right)}^2+{\left(-1-1\right)}^2}$
$\Rightarrow$radius = $\sqrt{4+4}$
$\Rightarrow$radius = $\sqrt{8}$
$\Rightarrow$radius = 2$\sqrt{2}$ units

Regards