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

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).

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anju2710 anju2710 · Answer 2 · 2017-09-11T10:57:54+02:00

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