This one is pretty tough. Frankly, I'm surprised you're having it assigned
in middle school.
If you sketch the points on paper, you'll see that there are two possibilities:
-- 'x' could be ' 4 ', so the third point would be at (0, 4). Then the right triangle
is upside down, with one of its legs up on top. Both legs are pieces of the 'x'
and 'y' axes, and the hypotenuse is the line between the origin and (-6, 4).
-- Another possibility: The third point could be up at y=4 and way over to the
right of the y-axis, and the right angle could be at the origin (0, 0). Here's how
to find it:
If the third point is over there, and the right angle is at the origin, then the line
from (-6, 4) to (0, 0) and the line from (x, 4) to (0, 0) are perpendicular.
You know how to find the slopes of these two lines.
- From (-6, 4) to (0, 0), the slope is (6 / -4) = - 3/2 .
- From (x, 4) to (0, 0), the slope is (-x / -4) = x/4 .
When two lines are perpendicular, their slopes are negative reciprocals ...
(one slope) = -1 / (the other slope).
So - 3/2 = -1 / (x/4)
Multiply each side by -1 : 3/2 = 1 / (x/4)
Multiply each side by x/4 : (3/2) (x/4) = 1
Multiply each side by 2/3 : x/4 = 2/3
Multiply each side by 4 : x = 8/3 = 2 and 2/3 .
