The hypotenuse of an isosceles right angled triangle has its ends at the

Question

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

My Attempt,

From the given information, I found the equation of the hypotenuse, using.. $$y-y_1=\frac {y_2-y_1}{x_2-x_1} (x+x_1)$$ $$y-3=\frac {1-3}{-4-1} (x-1)$$ $$y-3=\frac {-2}{-5} (x-1)$$ $$-5y+15=-2x+2$$ $$2x-5y+13=0$$,

...

Answer 1

If triangle has its vertices on a circle and one side is a diameter of the circle, then the angle opposite that side is a right angle.

The center of the circle is $C=\left(-\frac{3}{2},2\right)$ and the radius is $\frac{\sqrt{29}}{2}$ so its equation can be found. The slope of the perpendicular bisector of the segment $AB$ has slope $-\frac{5}{2}$ and contains $C$. The line intersects the circle at $E$ and at $D$ but all that is wanted is the slope of $EA$ and $BE$ which can be found by finding the coordinates of $E$. The slope of $DA$ is the same as $BE$ and the slope of $DB$ is the same as the slope of $EA$.

However, that way is considerably messy. It is probably easier to use trigonometry.

The line $AB$ makes an angle $\theta=\arctan\left(\frac{2}{5}\right)$ with the horizontal. The side $AD$ makes an angle $45^\circ+\arctan\left(\frac{2}{5}\right)$ so its slope is $\tan\left(45^\circ+\arctan\left(\frac{2}{5}\right)\right)=\frac{7}{3}$. Since $EA$ is perpendicular to $AD$ it has slope $-\frac{3}{7}$.

Note: This uses the identity $\tan(X+Y)=\dfrac{\tan X+\tan Y}{1-\tan X\tan Y}$.

The equation of the lines containing the four sides satisfying the conditions of the problem are

1. AB: $y-3=\frac{7}{3}(x-1)$
2. BD: $y-1=\frac{7}{3}(x+4)$
3. AD: $y-3=-\frac{3}{7}(x-1)$
4. BE: $y-1=-\frac{3}{7}(x+4)$

These simplify as follows:

1. AB: $\quad7x-3y+2=0$
2. BD: $\quad7x-3y+31=0$
3. AD: $\quad3x+7y-24=0$
4. BE: $\quad3x+7y+5=0$

Answer 2

Hint -

Let ABC be a right angle triangle. Right angled at B.

Let point A(1,3) and C(-4,1)

Then suppose slope of AB = m.

And AB.BC = -1

So BC = $ \frac{-1}{m}$

Find equations of AB and BC using slopes m and $ \frac{-1}{m}$ respectively.

Answer 3

Let third point be P(x,y)

Let A (1,3) and B(-4,1)

Since it's isosceles triangle PA=PB

${PA}^2 = {PB}^2$

$ (x-1)^2 + (y-3)^2 = (x+4)^2 + (y-1)^2 $

After solving gives $4y +10x +7=0 \text{equation①}$

Also ${PA}^2 + {PB}^2 = {AB}^2 $

$(x-1)^2 + (y-3)^2 + (x+4)^2 + (y-1)^2 = 5^2 + 2^2$

Which after solving gives $2x^2 + 2y^2 +6x - 8y=1 \text{equation ②}$

From equation ① $8y+ 20x + 14 =0$and $-8y=20x+14$ substituting in equation ②

Gives $x^2 + y^2 +13x +7 =0$ ( which is combined equation for legs )

Answer 4

The locus of $C$ are two semi-circles on opposite sides of $AB$.

There are many equation pairs, $C$ is a variable point on the semi-circles.

Mid-point coordinates are $ h=-3/2,k=2 $

Radius $\, R = \sqrt{(5/2)^2+ (2/2)^2 } = \sqrt{29}/2 $

$C$ the right triangle vertex. $AB$ is diameter/hypotenuse for circle with $(x,y)$ parametric equations

$$ x =h+ R \cos \theta ,\, y =k+ R \sin \theta ,\,$$

Can you now find equations of $CA,CB?$ connecting to $C$ to $AB?$