Question

Find the equation of the median and altitude of Triangle ABC through A where the vertices
are A(6,2), B(-5, -1) and C(1,9)

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

x + 4y - 14 = 0

5x - 3y  - 24 = 0

Step-by-step explanation:

Median drawn passing through the vertex A intersect the side BC at the midpoint.

D = $\frac{x_1 + y_1}{2}$       $\frac{x_1}{y_1}{2}$

D =  $\frac{-5 + 1}{2}$            $\frac{-1 + 9}{2}$

D = $\frax{-4}{2}$                   $\frac{8}{2}$

D = (-2, 4)

Equation of the median AD

$\frac{(y - y_1)}{y_2 - y_1}$ = $\frac{(x - x_1)}{x_2 - x_1}$

A (6, 2) D(-2, 4)

$\frac{(y - 2)}{4 - 2}$ = $\frac{(x - 6)}{-2 - 6}$

$\frac{(y - 2)}{2}$ = $\frac{(x - 6)}{-8}$

-8(y - 2) = 2(x - 6)

2x + 8y -12 - 16 = 0

2x + 8y -28 = 0

x + 4y - 14 = 0

If the line passing through the vertex A is altitude, then it will be perpendicular to BC.

Slope of BC

B(6,2) C(1, 9)

m = $\frac{y_2 - y_1}{x_2 - x_1}$

m = $\frac{9 + 1}{1 + 5}$

m =  $\frac{10}{6}$

m =  $\frac{5}{3}$

Equation of altitude passing through A.

$(y - y_1)$ = $m\times (x - x_1)$

A(6, 2) and m = $\frac{5}{3}$

$3\times (y - 2)$ = $5\times (x - 6)$

3y - 6 = 5x - 30

5x - 3y  - 24 = 0