Question

[If 3x -5y+k 0 is a median of the triangle ABC whose vertices are A(-1,3), B(0, 4) C(-5, 2) then the value of k is]

Answer 1

Solution:

Concept used:

The centroid of a triangle having vertices

$({x_1},{y_1}),({x_2},{y_2}),({x_3},{y_3})$ is

$(\frac{{x_1}+{x_2}+{x_3}}{3},\frac{{y_1}+{y_2}+{y_3}}{3})$

Given:

The vertice of triangle are A(-1,3), B(0, 4) C(-5, 2)

The centroid of triangle ABC is

$(\frac{{x_1}+{x_2}+{x_3}}{3},\frac{{y_1}+{y_2}+{y_3}}{3})$

$(\frac{-1+0-5}{3},\frac{3+4+2}{3})\\\\(\frac{-6}{3},\frac{9}{3})$

(-2,3)

since the median 3x -5y+k=0 passes through (-2,3),

3(-2)-5(3)+k=0

-6-15+k=0

k=21