Question

Find the incenter of the triangle formed by the straight lines y=√3x,y=-√3x
and y=3​

Answer 1

The incenter of thr triangle formed by the straight lines  $y=\sqrt{3} x,y=-\sqrt{3}x,y=3$ is $(2,0)$

• Given,
• equations of straight lines: $y=\sqrt{3} x,y=-\sqrt{3}x,y=3$
• The triangle formed by the intersection of theses lines have vertices, given by,
• $A(0,0),B(\sqrt{3},3),C(-\sqrt{3},3)$
• Now,
• the lengths of sides of triangle formed are given by,

• $AB=\sqrt{(\sqrt 3-0)^2+(3-0)^2} =\sqrt{3+9} =2\sqrt{3} \\BC=\sqrt{(-\sqrt 3-\sqrt 3)^2+(3-3)^2} =\sqrt{(2\sqrt 3)^2} =2\sqrt{3} \\AC=\sqrt{(-\sqrt 3-0)^2+(3-0)^2} =\sqrt{3+9} =2\sqrt{3}$

• we use the below formula to find the incenter of triangle formed,

• $I=(\dfrac{ax_1+bx_2+cx_3}{a+b+c}, \dfrac{ay_1+by_2+cy_3}{a+b+c} ) \\\\=(\dfrac{2\sqrt 3(0)+2\sqrt 3(\sqrt 3)+2\sqrt 3(-\sqrt 3)}{2\sqrt 3+2\sqrt 3+2\sqrt 3}, \dfrac{2\sqrt 3(0)+2\sqrt 3(3)+2\sqrt 3(3)}{2\sqrt 3+2\sqrt 3+2\sqrt 3} )\\\\=(\dfrac{0+6-6}{6 \sqrt 3},\dfrac{0+6\sqrt 3+6\sqrt 3}{6 \sqrt 3})\\\\=(0,2)$

Answer 2

Step-by-step explanation:

........check itttttt