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

...

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    Note that there will be two solutions because the legs can be on either side of the hypotenuse. – 2017-01-29
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    @mary, I know that. so its legs, not leg? . – 2017-01-29
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    Right, and there are two possible positions for the vertex. This usually means that this will result in a quadratic equation. – 2017-01-29
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    @marty, how do I get that? – 2017-01-29
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    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. – 2017-01-29
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    @John Wayland Bales, I.could not get that. Could you please elaborate? – 2017-01-29
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    You have the endpoints of the hypotenuse and that it is a right isosceles triangle. So the vertex opposite the hypotenuse must lie on the perpendicular bisector of the hypotenuse as will the vertices of all isosceles triangles. Since the hypotenuse will be a diameter of a circle drawn at its midpoint and passing through its endpoints, every triangle having its other vertex on that circle will form a right triangle. – 2017-01-29
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    @John Wayland, How do I solve further to get the equation – 2017-01-29

4 Answers 4

4

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

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    @@John Wayland, What are the perpendicular sides whose equations are to be calculated? – 2017-01-29
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    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? – 2017-01-29
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    Wayland, Please add it. It seems quite confusing for me. – 2017-01-29
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    Did you understand how I use the tangents to find the slope of the sides? Have you studied trigonometry? – 2017-01-29
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    Wayland, I understand Trigonometry and I am well familiar with this topic (Trigonometry), too. – 2017-01-29
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    So we can forget about perpendicular bisectors, circles, etc, except it allows us to make a nice diagram. The actual solution begins after the sentence "However, that way is considerably messy." – 2017-01-29
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    @JohnWaylandBales I really like your sketch. What software did you use to generate it? – 2017-01-29
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    @@John Wayland, With which horizontal, does $AB$ make an angle of $\theta $? – 2017-01-29
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    If line $AB$ be extended it intersects every horizontal line at an angle $\theta=\arctan\left(\frac{2}{5}\right)$ because the slope of the line $AB$ is $\frac{2}{5}$. – 2017-01-29
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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.

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    @@Kanwaljit Singh, I need your help with another question, too. Will you please help ? – 2017-01-29
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    I will try. Send me link. – 2017-01-29
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    See this for complete and easy solution. http://teachoo.com/2695/628/Misc-17---Hypotenuse-of-a-right-angled-triangle-has-its-ends-at/category/Miscellaneous/ – 2017-01-29
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    @@Kanwaljit Singh, math.stackexchange.com/questions/2117470/question-on-graph-and-statistics Please check the link above. – 2017-01-29
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    @@Kanwaljit Singh, Did you see the question? – 2017-01-29
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    Yes I am looking. – 2017-01-29
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    @@Kanwaljit Singh, Any idea about the question? – 2017-01-29
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    @@Kanwaljit, Did you get the answer for the graph? – 2017-01-29
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    Actually your graph shows frequency and I saw graphs questions with always cumulative frequency. So I have confusion. – 2017-01-29
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    @@Kanwaljit, That's the graph for cumulative frequency. – 2017-01-29
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    John getting answer as your book answer. So what doubt you still have in this question? – 2017-01-29
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    I have not understood some of the steps. So, I am waiting for his reply... – 2017-01-29
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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 $AB?$

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    @@Narasimham, What are $A,B,C$? Could you please show ? – 2017-01-29
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    @@Narsimham, Why do you have $C$ on the line $AB$? – 2017-01-29
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    @@Narasimham, could you please explain me about the circle with parametric equation? – 2017-01-29
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    Do you know circle equation in parametric form? The above is an example. – 2017-01-29
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    @@Narasimham, No i don't know. So, please elaborate. – 2017-01-29
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    Please learn from textbook, but I shall add elaborative labels. – 2017-01-29
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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 )

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    The second condition gives a quadratic, so there are two solutions. – 2017-01-29
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    @martycohen yes :-) hope OP don't mess while solving – 2017-01-29
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    It would be nice if you did the second part too. – 2017-01-29
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    @martycohen OK,I am adding – 2017-01-29
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    @Ramanujan, Actually I have two equations as answer in my book, $7x-3y+31=0, 3x+7y=24$. – 2017-01-29
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    I think you have some mistake how from equation (1), -8y=20x+7y? – 2017-01-29
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    @KanwaljitSingh I am fixing it , thanks for figuring out :) – 2017-01-29
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    Your welcome... – 2017-01-29
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    @KanwaljitSingh now is it correct? ( But still it doesn't matches with oP answer :( ) – 2017-01-29
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    Your fix is now ok. But I see your answer is not matching. – 2017-01-29
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    Try to solve it in different ways. I also trying. – 2017-01-29
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    @@@Kanwaljit Singh, Did you see John Wayland's answers. That matches with the answers in my book – 2017-01-29