Question

Please help! "Indicate the equation of the given line in standard form. The line containing the midpoints of the legs of right triangle ABC where A(-5, 5), B(1, 1), and C(3, 4) are the vertices."

Answer 1

Nice. We're not really told which is the right angle. We can tell from the squared distances:

$AB^2=(1 - -5)^2 + (1-5)^2=6^2+4^2=52$

$BC^2=(3-1)^2+(4-1)^2=2^2+3^2=13$

$AC^2=(3- -5)^2+(4 -5)^2=8^2+1^2=65$

We see $AC^2=AB^2+BC^2$ so B is the right angle.

That was preliminary and probably indicated in an associated figure.

Now we want the line through the midpoint of the legs, AB and BC, in standard form.

The midpoint of AB is the average of the coordinates of A and B:

$\left( \dfrac{-5 +1}{2}, \dfrac{5+1}{2} \right) = (-2,3)$

The midpoint of BC is

$\left( \dfrac{1 + 3}{2}, \dfrac{1 + 4}{2} \right) = ( 2, \frac 5 2)$

The line through those points is

$(y - 3)(2 - -2)=(x - -2)(\frac 5 2 - 3)$

$4y - 12 = (x+2)(- \frac 1 2)$

$8y - 24 = -x -2$

$x + 8y = 22$

That's the answer. Let's check it.

Midpoint of AB is (-2,3), -2 + 8(3)=22 good.

Midpoint of BC is (2,5/2), 2+8(5/2)=22 good.

$\textrm{Answer: } x + 8y = 22$