0
$\begingroup$

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.. $$y-y_1=\frac {y_2-y_1}{x_2-x_1} (x+x_1)$$ $$y-3=\frac {1-3}{-4-1} (x-1)$$ $$y-3=\frac {-2}{-5} (x-1)$$ $$-5y+15=-2x+2$$ $$2x-5y+13=0$$,

...

| cite | improve this question | | | | |
$\endgroup$
  • $\begingroup$ Note that there will be two solutions because the legs can be on either side of the hypotenuse. $\endgroup$ – marty cohen Jan 29 '17 at 5:25
  • $\begingroup$ @mary, I know that. so its legs, not leg? . $\endgroup$ – pi-π Jan 29 '17 at 5:27
  • $\begingroup$ Right, and there are two possible positions for the vertex. This usually means that this will result in a quadratic equation. $\endgroup$ – marty cohen Jan 29 '17 at 5:30
  • $\begingroup$ @marty, how do I get that? $\endgroup$ – pi-π Jan 29 '17 at 5:32
  • $\begingroup$ The vertex will lie on the perpendicular bisector of the hypotenuse and on the circle with center at the midpoint and containing the endpoints of the hypotenuse. That will give you the two solutions. $\endgroup$ – John Wayland Bales Jan 29 '17 at 5:34
4
$\begingroup$

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: $\quad3x+7y+5=0$

Right Triangle Problem

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ @@John Wayland, What are the perpendicular sides whose equations are to be calculated? $\endgroup$ – pi-π Jan 29 '17 at 6:43
  • $\begingroup$ I had to go back and read the question. I though you only wanted the slopes of the sides. If you want the equations, then you need four equations for the lines containing EA, BD, AD and EB since all four satisfy the conditions. But now that you have the slope of all four and know that each of the four contain either the point A or the point B you just use the point-slope equation to find the four equations. Can you do that or do in need to add it to the solution I gave? $\endgroup$ – John Wayland Bales Jan 29 '17 at 6:51
  • $\begingroup$ Wayland, Please add it. It seems quite confusing for me. $\endgroup$ – pi-π Jan 29 '17 at 6:53
  • $\begingroup$ Did you understand how I use the tangents to find the slope of the sides? Have you studied trigonometry? $\endgroup$ – John Wayland Bales Jan 29 '17 at 6:54
  • $\begingroup$ Wayland, I understand Trigonometry and I am well familiar with this topic (Trigonometry), too. $\endgroup$ – pi-π Jan 29 '17 at 6:58
1
$\begingroup$

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.

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ @@Kanwaljit Singh, I need your help with another question, too. Will you please help ? $\endgroup$ – pi-π Jan 29 '17 at 5:42
  • $\begingroup$ I will try. Send me link. $\endgroup$ – Kanwaljit Singh Jan 29 '17 at 5:44
  • $\begingroup$ See this for complete and easy solution. teachoo.com/2695/628/… $\endgroup$ – Kanwaljit Singh Jan 29 '17 at 5:45
  • $\begingroup$ @@Kanwaljit Singh, math.stackexchange.com/questions/2117470/question-on-graph-and-statistics Please check the link above. $\endgroup$ – pi-π Jan 29 '17 at 5:46
  • $\begingroup$ @@Kanwaljit Singh, Did you see the question? $\endgroup$ – pi-π Jan 29 '17 at 5:56
1
$\begingroup$

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.

Lines in 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 $AB?$

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ @@Narasimham, What are $A,B,C$? Could you please show ? $\endgroup$ – pi-π Jan 29 '17 at 7:10
  • $\begingroup$ @@Narsimham, Why do you have $C$ on the line $AB$? $\endgroup$ – pi-π Jan 29 '17 at 7:19
  • $\begingroup$ @@Narasimham, could you please explain me about the circle with parametric equation? $\endgroup$ – pi-π Jan 29 '17 at 7:25
  • $\begingroup$ Do you know circle equation in parametric form? The above is an example. $\endgroup$ – Narasimham Jan 29 '17 at 7:30
  • $\begingroup$ @@Narasimham, No i don't know. So, please elaborate. $\endgroup$ – pi-π Jan 29 '17 at 7:32
0
$\begingroup$

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 )

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ The second condition gives a quadratic, so there are two solutions. $\endgroup$ – marty cohen Jan 29 '17 at 5:42
  • $\begingroup$ @martycohen yes :-) hope OP don't mess while solving $\endgroup$ – Fawad Jan 29 '17 at 5:43
  • $\begingroup$ It would be nice if you did the second part too. $\endgroup$ – marty cohen Jan 29 '17 at 5:44
  • $\begingroup$ @martycohen OK,I am adding $\endgroup$ – Fawad Jan 29 '17 at 5:46
  • $\begingroup$ @Ramanujan, Actually I have two equations as answer in my book, $7x-3y+31=0, 3x+7y=24$. $\endgroup$ – pi-π Jan 29 '17 at 6:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.