Let A(0,0), B(3,0), C(0,-4) be the vertices o | vedclass

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

Question

(B) The vertices are $A(0,0)$,$B(3,0)$,and $C(0,-4)$.Calculate the side lengths:$c = AB = \sqrt{(3-0)^2 + (0-0)^2} = \sqrt{9} = 3$$b = AC = \sqrt{(0-0

Vedclass · Accepted Answer

$(1,-1)$

Vedclass · Answer

(B) The vertices are $A(0,0)$,$B(3,0)$,and $C(0,-4)$.Calculate the side lengths:$c = AB = \sqrt{(3-0)^2 + (0-0)^2} = \sqrt{9} = 3$$b = AC = \sqrt{(0-0)^2 + (-4-0)^2} = \sqrt{16} = 4$$a = BC = \sqrt{(0-3)^2 + (-4-0)^2} = \sqrt{9+16} = \sqrt{25} = 5$The coordinates of the incentre $(I)$ are given by the formula:$I = \left( \frac{ax_1 + bx_2 + cx_3}{a+b+c}, \frac{ay_1 + by_2 + cy_3}{a+b+c} \right)$Here,$(x_1, y_1) = (0,0)$,$(x_2, y_2) = (3,0)$,and $(x_3, y_3) = (0,-4)$.$I = \left( \frac{5(0) + 4(3) + 3(0)}{5+4+3}, \frac{5(0) + 4(0) + 3(-4)}{5+4+3} \right)$$I = \left( \frac{12}{12}, \frac{-12}{12} \right) = (1, -1)$.

Let $A(0,0), B(3,0), C(0,-4)$ be the vertices of $\triangle ABC$. Then the coordinates of the incentre of $\triangle ABC$ are:

Similar Questions

The incentre of the triangle formed by the straight lines $y=\sqrt{3}x$,$y=-\sqrt{3}x$ and $y=3$ is

$A(1, -2), B(-2, 3), C(-1, -3)$ are the vertices of a triangle $ABC$. $L_1$ is the perpendicular drawn from $A$ to $BC$ and $L_2$ is the perpendicular bisector of $AB$. If $(l, m)$ is the point of intersection of $L_1$ and $L_2$, then $26m - 3 =$ (in $l$)

The orthocentre of the triangle formed by the lines $xy = 0$ and $x + y = 1$ is

Find the orthocenter of the triangle whose vertices are $(0, 0), (2, -1),$ and $(1, 3).$

For a triangle formed by the vertices $(0,0)$,$(4,0)$,and $(3,4)$,the orthocenter is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module