Question

Triangle PQR has vertices P(4,-1),Q(-2,7), and R(9,9). Find an equation of the median and altitude from R.

Answer 1

If we are drawing altitude from R the PQ will be base of this triangle. So we can say Median from R will cut on Midpoint of PQ. So firstly we will find mid point of PQ Which is S given as below:
P(4,-1) , Q (-2,7)

Let Co-ordinate of S is (x,y)
So $x = \frac{x_1+x_2}{2}$
$x = \frac{4-2}{2} = \frac{2}{2} = 1$

$y = \frac{y_1+y_2}{2}$
$y = \frac{-1+7}{2} = \frac{6}{2}= 3$
So midpoint of PQ is S(1,3).

Now we can find equation of RS
Co-ordinate of R is (9,9) and and S is (1,3).

In general equation of line from two point $(x_1 , y_1)$ and $(x_2,y_2)$ is
$(y-y_1) = \frac{y_2-y_1}{x__2-x_1}(x-x_1)$

So equation of RS is

$(y-9) = \frac{9-3}{9-1} (x-1)$
$(y-9) = \frac{6}{8}(x-1)$
$8(y-9) = 6(x-1)$
$8y-72 = 6x - 6$
$-72+6 = 6x - 8y$
$-66 = 6x-8y$

So the equation of median is 6x - 8y = -66

Answer 2

Both of the equation of the median and altitude from R are the same: y=(3/4)x + 9/4, this means that they are at equal length so the triangle PQR is an isosceles (a triangle that has two sides of the same length).