If the centroid of Delta ABC is G(3, 3) and t | vedclass

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

Question

(A) Let the coordinates of vertex $A$ be $(x, y)$.The centroid $G(x_g, y_g)$ of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given

Vedclass · Accepted Answer

$\frac{69}{2}$

Vedclass · Answer

(A) Let the coordinates of vertex $A$ be $(x, y)$.The centroid $G(x_g, y_g)$ of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $G = (\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3})$.Given $G(3, 3)$,$B(-1, 2)$,and $C(2, -3)$,we have:$3 = \frac{x - 1 + 2}{3} \implies 9 = x + 1 \implies x = 8$.$3 = \frac{y + 2 - 3}{3} \implies 9 = y - 1 \implies y = 10$.So,vertex $A$ is $(8, 10)$.The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.Area $= \frac{1}{2} |8(2 - (-3)) + (-1)(-3 - 10) + 2(10 - 2)|$.Area $= \frac{1}{2} |8(5) + (-1)(-13) + 2(8)|$.Area $= \frac{1}{2} |40 + 13 + 16| = \frac{1}{2} |69| = \frac{69}{2}$ square units.

If the centroid of $\Delta ABC$ is $G(3, 3)$ and the vertices are $B(-1, 2)$ and $C(2, -3)$,then find the area of $\Delta ABC$.

Similar Questions

Find a relation between $x$ and $y$ if the points $(x, y), (1, 2)$ and $(7, 0)$ are collinear.

Prove that the area of the triangle formed by joining the midpoints of the sides of a triangle is one-fourth the area of the original triangle.

In $\Delta ABC$,if $A(1, 7)$,$B(2, 4)$,and $C(k, 5)$ and $\angle A$ is a right angle,then find the value of $k$.

Given points $A(3, 4)$ and $B(5, -2)$. Find the point $P(x, y)$ on the plane such that $PA = PB$ and the area of $\Delta PAB = 10$.

$A (h, k), B (1, 1)$ and $C (2, 1)$ are the vertices of $\Delta ABC$. If the area of $\Delta ABC$ is $1$,then find the possible values of $k$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module