JoshVD
contestada

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

Respuesta :

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).
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 [tex]x = \frac{x_1+x_2}{2} [/tex]
[tex]x = \frac{4-2}{2} = \frac{2}{2} = 1 [/tex]

[tex]y = \frac{y_1+y_2}{2} [/tex]
[tex]y = \frac{-1+7}{2} = \frac{6}{2}= 3 [/tex]
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 [tex](x_1 , y_1)[/tex] and [tex](x_2,y_2)[/tex] is
[tex](y-y_1) = \frac{y_2-y_1}{x__2-x_1}(x-x_1) [/tex]

So equation of RS is

[tex](y-9) = \frac{9-3}{9-1} (x-1)[/tex]
[tex](y-9) = \frac{6}{8}(x-1) [/tex]
[tex]8(y-9) = 6(x-1)[/tex]
[tex]8y-72 = 6x - 6[/tex]
[tex]-72+6 = 6x - 8y[/tex]
[tex]-66 = 6x-8y[/tex]

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