The hypotenuse of an isosceles...

Question

The hypotenuse of an isosceles right angled triangle has its ends at the points $(1,3)$ and $(-4,1)$. Find the equation of the legs (perpendicular sides) of the triangle.

My Approach,

Since the triangle us an isosceles right angles triangle, it has three angles $90°$, $45°$ and $45°$ respectively.

Also, the equation of the hypotenuse is given by: $$y-y_1=\frac {y_2-y_1}{x_2-x_1} (x-x_1)$$ $$y-3=\frac {1-3}{-4-1} (x-1)$$ $$y-3=\frac {2}{5} (x-1)$$ $$5y-15=2x-2$$ $$2x-5y+13=0$$

I am stuck at here. Please help me to complete it.

Answer 1

Because you are given two points where the hypotenuse starts and ends, the first thing you should do is find the slope: $$ m = \frac{y_2-y_1}{x_2-x_1} = \frac{3-1}{1-(-4)} = \frac{2}{5} $$ With the generic linear equation form $y = mx+b$, plug in $m$, which was derived in the previous step. So, the equation is now $y = \frac{2}{5}x + b$.

In order to find $b$, plug in $(x,y)$ that satisfies the equation. You can choose one of the points of the hypotenuse because one of those two points obviously lies on the line.

$$ 1 = \frac{2}{5}(-4) + b \\ 1 = \frac{-8}{5} + b \\ b = \frac{13}{5} $$

Plug $b$ back into the equation, giving you $y = \frac{2}{5}x + \frac{13}{5}$. For a sanity check, I plotted the two points and the line that goes through it – it's a perfect fit!

Answer 2

For an isosceles right angled triangle, the perpendicular from the vertex opposite the hypotenuse to the hypotenuse bisects the hypotenuse. Moreover, its length is equal to half the length of the hypotenuse.

So, it is easy to find the third vertex. By the mid-point formula, the mid-point of the hypotenuse is, $$x_m=\frac{1-4}{2}=-\frac{3}{2}$$ $$y_m=\frac{3+1}{2}=2$$

Now, find the distance between this and any end point. The length of the perpendicular is the same distance. Moreover, the slope of the perpendicular is the negative of the reciprocal of the slope of the hypotenuse, a property of perpendicular lines. So, the equation of the perpendicular can be found and the third vertex can be found.

Using the two point form of the equation of a line, the equations of the legs can also be found.