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:

Equation of Median through A:

x + 4y - 14 = 0

Equation of Altitude through A:

3x + 5y - 28 = 0

Step-by-step explanation:

Vertex of ∆ABC are A(6,2), B(-5, -1) and C(1,9)

Since through A ,median bisect the side BC.

So first find the mid point of BC,let that point is D

Coordinates of D(x,y)

$= \frac{ - 5 + 1}{2}, \frac{ - 1 + 9}{2} \\ \\ = ( - 2 ,\: 4)$

Equation of a line passes through two points

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

Equation of line AD,thus median AD

$y -2 = \bigg(\frac{4 - 2}{ - 2 - 6}\bigg) x - 6\\ \\y - 2 = \frac{2}{ - 8} (x - 6) \\ \\ y - 2 = \frac{ - 1}{4} (x - 6) \\ \\ 4y - 8 = - x + 6 \\ \\ x + 4y - 14 = 0$

is the equation of median.

Now to find the equation of altitude,as altitudes is perpendicular to the BC,so find the Slope of BC first

$slope \: of \: BC = \bigg( \frac{ - 1 - 9}{ - 5 - 1} \bigg) \\ \\ = \frac{ - 10}{ - 6} \\ \\ slope \: of \: BC= \frac{5}{3}$

Slope of altitude -3/5

Altitude meet at point A,

Thus equation of Altitude

$y - 2 = \frac{ - 3}{5} (x - 6) \\ \\ 5y - 10 = - 3x + 18 \\ \\ 3x + 5y - 28 = 0$

Hope it helps you.