Question

If A(-1, 8), B(4, -2) and C(-5, -3) are the vertices of a triangle, then the equation of the median through (-1, 8) will be .

• A. 21x + y + 13 = 0
• B. 2x + y - 6 = 0
• C. 11x - 4y + 43 = 0

Answer 1

The correct option is A 21x + y + 13 = 0

Let's assume median which passes through A cuts the line BC at D.

D is the midpoint of line BC

D(x,y) = $(\frac{4-5}{2},\frac{-2-3}{2})=(\frac{-1}{2},\frac{-5}{2})$

Equation of the line passing through (-1,8) and $(\frac{-1}{2},\frac{-5}{2})$

Using the formula
$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1),weget$

$\Rightarrow y-8=\frac{\frac{-5}{2}-8}{\frac{-1}{2}+1}(x+1)$

$\Rightarrow y-8=\frac{\frac{-5-16}{2}}{\frac{-1+2}{2}}(x+1)$

$\Rightarrow y-8=\frac{-21}{1}(x+1)$

$\Rightarrow y-8=-21x-21$

$\Rightarrow 21x+y+13=0$

Equation of the line (median) that passes through (-1,8) is
$21x+y+13=0$