If (0, 0),(6, 0),and (6, 8) are the vertices | vedclass

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

Question

(D) Let the vertices be $A(0, 0)$,$B(6, 0)$,and $C(6, 8)$.Calculate the side lengths:$c = AB = \sqrt{(6-0)^2 + (0-0)^2} = 6$$a = BC = \sqrt{(6-6)^2 +

Vedclass · Accepted Answer

$(4, 2)$

Vedclass · Answer

(D) Let the vertices be $A(0, 0)$,$B(6, 0)$,and $C(6, 8)$.Calculate the side lengths:$c = AB = \sqrt{(6-0)^2 + (0-0)^2} = 6$$a = BC = \sqrt{(6-6)^2 + (8-0)^2} = 8$$b = AC = \sqrt{(6-0)^2 + (8-0)^2} = \sqrt{36+64} = 10$The formula for the incentre $(I)$ is $\left( \frac{ax_1 + bx_2 + cx_3}{a+b+c}, \frac{ay_1 + by_2 + cy_3}{a+b+c} \right)$.Substituting the values:$I = \left( \frac{8(0) + 10(6) + 6(6)}{8+10+6}, \frac{8(0) + 10(0) + 6(8)}{8+10+6} \right)$$I = \left( \frac{0 + 60 + 36}{24}, \frac{0 + 0 + 48}{24} \right)$$I = \left( \frac{96}{24}, \frac{48}{24} \right) = (4, 2)$.

If $(0, 0)$,$(6, 0)$,and $(6, 8)$ are the vertices of a triangle,what is its incentre?

Similar Questions

Consider a $\triangle PQR$ in which the relation $QR^2 + PR^2 = 5PQ^2$ holds. Let $G$ be the point of intersection of medians $PM$ and $QN$. Then,$\angle QGM$ is always

If $(\alpha, \beta)$ is the orthocentre of the triangle $ABC$ with vertices $A(3, -7)$,$B(-1, 2)$,and $C(4, 5)$,then $9\alpha - 6\beta + 60$ is equal to:

If $A(x_1, y_1)$,$B(x_2, y_2)$,and $C(x_3, y_3)$ are the vertices of a triangle with side lengths $a, b, c$ opposite to vertices $A, B, C$ respectively,then the excentre with respect to vertex $B$ is:

The coordinates of the orthocentre of the triangle bounded by the lines $4x - 7y + 10 = 0$,$x + y = 5$,and $7x + 4y = 15$ are:

The incentre of the triangle formed by the vertices $(0, 0)$,$(5, 12)$,and $(16, 12)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module