The sides of a triangle A B C lie on the lines 3 x+4 y=0, 4 x+3 y=0, and x=3. Let (h, k) be the center of the circle inscribed in triangle A B C. The value of (h+k) equals.

Question

The sides of a triangle A B C lie on the lines 3 x+4 y=0, 4 x+3 y=0, and x=3. Let (h, k) be the center of the circle inscribed in triangle A B C. The value of (h+k) equals. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

The equation of the angle bisector of angle A is (3 x+4 y/5)=± (4 x+3 y/5) text or x=± y <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/94a6a5a06333620e41b239ac7ef7997f-.png" /> The equation of internal bisector is x=-y Since h and k lie on the line x=-y, we have h+k=0

Q. The sides of a triangle $A BC$ lie on the lines $3 x + 4 y = 0$ , $4 x + 3 y = 0$ , and $x = 3$ . Let $(h, k)$ be the center of the circle inscribed in $△ A BC$ . The value of $(h + k)$ equals.

The equation of the angle bisector of angle $A$ is
$\frac{3 x + 4 y}{5} = \pm \frac{4 x + 3 y}{5} or x = \pm y$

The equation of internal bisector is $x = - y$ Since $h$ and $k$ lie on the line $x = - y$ , we have
$h + k = 0$

Q. The sides of a triangle ABC lie on the lines 3x+4y=0, 4x+3y=0, and x=3. Let (h,k) be the center of the circle inscribed in △ABC. The value of (h+k) equals.

The equation of the angle bisector of angle A is 53x+4y​=±54x+3y​ or x=±y The equation of internal bisector is x=−y Since h and k lie on the line x=−y, we have h+k=0

Q. The sides of a triangle $A BC$ lie on the lines $3 x + 4 y = 0$ , $4 x + 3 y = 0$ , and $x = 3$ . Let $(h, k)$ be the center of the circle inscribed in $△ A BC$ . The value of $(h + k)$ equals.

The equation of the angle bisector of angle $A$ is
$\frac{3 x + 4 y}{5} = \pm \frac{4 x + 3 y}{5} or x = \pm y$

The equation of internal bisector is $x = - y$ Since $h$ and $k$ lie on the line $x = - y$ , we have
$h + k = 0$