Find the incenter of the triangle formed by t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The vertices of the triangle formed by the axes and the line $\frac{x}{a} + \frac{y}{b} = 1$ are $O(0, 0)$,$A(a, 0)$,and $B(0, b)$.Let the side le

Vedclass · Accepted Answer

$\left( \frac{ab}{a + b + \sqrt{a^2 + b^2}}, \frac{ab}{a + b + \sqrt{a^2 + b^2}} \right)$

Vedclass · Answer

(B) The vertices of the triangle formed by the axes and the line $\frac{x}{a} + \frac{y}{b} = 1$ are $O(0, 0)$,$A(a, 0)$,and $B(0, b)$.Let the side lengths be $c = AB = \sqrt{a^2 + b^2}$,$b' = OB = b$,and $a' = OA = a$.The coordinates of the incenter $(I_x, I_y)$ are given by the formula $\left( \frac{a'x_1 + b'x_2 + c'x_3}{a' + b' + c'}, \frac{a'y_1 + b'y_2 + c'y_3}{a' + b' + c'} \right)$.Substituting the vertices $(0, 0)$,$(a, 0)$,and $(0, b)$ and side lengths $a' = b$,$b' = a$,and $c' = \sqrt{a^2 + b^2}$:$I_x = \frac{b(0) + a(a) + \sqrt{a^2 + b^2}(0)}{a + b + \sqrt{a^2 + b^2}} = \frac{ab}{a + b + \sqrt{a^2 + b^2}}$$I_y = \frac{b(0) + a(0) + \sqrt{a^2 + b^2}(b)}{a + b + \sqrt{a^2 + b^2}} = \frac{ab}{a + b + \sqrt{a^2 + b^2}}$Thus,the incenter is $\left( \frac{ab}{a + b + \sqrt{a^2 + b^2}}, \frac{ab}{a + b + \sqrt{a^2 + b^2}} \right)$.

Find the incenter of the triangle formed by the axes and the line $\frac{x}{a} + \frac{y}{b} = 1$.

Similar Questions

If the lines $3x + y - 4 = 0$,$x - ay - 10 = 0$,and $bx + 2y + 9 = 0$ form three successive sides of a rectangle in that order and the fourth side passes through $(1, 2)$,then the area of that rectangle (in sq. units) is

The sides of a triangle are $3x+2y-6=0$,$2x-3y+6=0$,and $x+2y+2=0$. If $P(0, b)$ lies either on the triangle or inside the triangle,then $b$ lies in the interval

What is the angle between the diagonals of the parallelogram formed by the lines $ℓx + my + n = 0$,$ℓx + my + n' = 0$,$mx + ℓy + n = 0$,and $mx + ℓy + n' = 0$?

What is the ratio of the area of $\triangle PQS$ to the area of $\triangle PQR$?

The equations of the perpendicular bisectors of the sides $AB$ and $AC$ of $\triangle ABC$ are $x-y+5=0$ and $x+2y=0$ respectively. If the coordinates of $A$ are $(1,-2)$,then the equation of the line $BC$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module