# Triangle problem in coordinate geometry

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• Nov 16th 2013, 06:00 PM
AaPa
Triangle problem in coordinate geometry
I have a triangle with two vertices known (1,3) and (-4,1).

I know a formula relating the acute angle formed by two lines and their slopes. It is

$tanA=\lvert\frac{m1-m2}{1+m1m2}\rvert$

I used this formula for the hypotenuse and one of the sides(m1) and then again used this formula for the hypotenuse and the other side(m2). I used the result m1m2=-1 in the second result and got two values of m1. But when I equate those values I get tanA=0.
What is wrong?
• Nov 16th 2013, 07:24 PM
ibdutt
Re: Triangle problem in coordinate geometry
What do you want to find out, for drawing a triangle we need to know three elements?
• Nov 16th 2013, 07:29 PM
AaPa
Re: Triangle problem in coordinate geometry
I don't want to find anything. I am just not able to understand why I am getting tanA=0 when it could be anything.
• Nov 16th 2013, 08:27 PM
ibdutt
Re: Triangle problem in coordinate geometry
Please show your work that will make this clear. In fact where are the slopes of tow other sides of the triangle? I Sri think some information is missing. With only this much information we gat infinitely many triangles and hence the problem?
• Nov 16th 2013, 08:58 PM
AaPa
Re: Triangle problem in coordinate geometry
there is no other information. I don't know what I am doing wrong with the information that is provided.
slope of hypotenuse=2/5

putting this in the formula for tan A.We get

$m1=\frac{2-5tanA}{2tanA+5}$

Similarly, I found m2 in terms of tan A and replaced m2 with -1/m1 to get
$m1=\frac{2+5tanA}{5-2tanA}$

On equating these two values of tanA I get tanA=0.
• Nov 16th 2013, 09:49 PM
SlipEternal
Re: Triangle problem in coordinate geometry
$m_1$ appears to be the slope of a vertical line. But, if it is not, then you can calculate it using $(x,y)$. $m_1 = \dfrac{3-y}{1-x}$ (note: this slope is not defined if $x=1$).

Then $m_2$ appears to be the slope of a horizontal line. Still, we can calculate it: $m_2 = \dfrac{1-y}{-4-x}$ (note: this slope is not defined if $x = -4$).

Then, you would have $\tan A = \left| \dfrac{\tfrac{3-y}{1-x} - \tfrac{1-y}{-4-x}}{1+\tfrac{3-y}{1-x}\tfrac{1-y}{-4-x}} \right| = \left| \dfrac{(4+x)(3-y)+(1-y)(1-x)}{(4+x)(x-1)+(1-y)(3-y)} \right|$

Simplifying, you get $\tan A = \left|\dfrac{13 + 2x -5y}{x^2+y^2+3x-4y-1} \right|$

Is that what you were looking for?
• Nov 16th 2013, 09:59 PM
AaPa
Re: Triangle problem in coordinate geometry
I was not looking for anything. I am just getting a wrong result that is tanA=0 but I am not able to find mistake in my work.
• Nov 16th 2013, 10:02 PM
SlipEternal
Re: Triangle problem in coordinate geometry
Oh, I misread the problem. This may not be the solution. The calculation I gave was for the angle between the lines with slopes $m_1,m_2$, not angle $A$. Also, if you are suggesting that the two lines are perpendicular, then $\dfrac{3-y}{1-x} = -\dfrac{-4-x}{1-y}$

So, possibly this would be helpful $\tan A = \left|\dfrac{\tfrac{3-y}{1-x} - \tfrac{2}{5}}{1+\tfrac{3-y}{1-x}\tfrac{2}{5}} \right| = \dfrac{\text{opp}}{\text{adj}} = \sqrt{\dfrac{(x+4)^2 + (y-1)^2}{(1-x)^2 + (3-y)^2}}$ The second equation just came from finding the length of each segment.

Then, simplifying the LHS and squaring both sides, you have:

$\left(\dfrac{13-5y+2x}{11-5x-2y}\right)^2 = \dfrac{(x+4)^2 + (y-1)^2}{(x-1)^2 + (3-y)^2}$

Apparently, this has solutions at $x=-4, y=3$ over the integers, and over the reals, it has solutions $x \approx -4.19$ or $x \approx 1.19$ and $y \approx 2$.
• Nov 16th 2013, 10:20 PM
AaPa
Re: Triangle problem in coordinate geometry
How did you get the solutions. It has infinite solutions.
• Nov 16th 2013, 10:27 PM
SlipEternal
Re: Triangle problem in coordinate geometry
Quote:

Originally Posted by AaPa
How did you get the solutions. It has infinite solutions.

A system of two equations in two variables may be solvable. This one is.

If $m_1$ and $m_2$ are the slopes of perpendicular lines, then $m_1 = -\dfrac{1}{m_2}$. So you have $\dfrac{3-y}{1-x} = -\dfrac{-4-x}{1-y}$

From calculating $\tan A$ in two different ways: $\left(\dfrac{13-5y+2x}{11-5x-2y}\right)^2 = \dfrac{(x+4)^2 + (y-1)^2}{(x-1)^2 + (3-y)^2}$

Then, I plugged those two into Wolframalpha, and it gave me the solutions.
• Nov 16th 2013, 10:38 PM
AaPa
Re: Triangle problem in coordinate geometry
But all that was given was a line segment with an arbitrary point(x,y). It could have lied at any point on the circle with the line segment as the diameter.
• Nov 16th 2013, 10:40 PM
SlipEternal
Re: Triangle problem in coordinate geometry
That's a good point. I dunno. Weird that Wolframalpha gave solutions...
• Nov 16th 2013, 10:44 PM
AaPa
Re: Triangle problem in coordinate geometry
please tell me my mistake in post 5.
• Nov 16th 2013, 10:53 PM
SlipEternal
Re: Triangle problem in coordinate geometry
The line with slope $m_2$ is not adjacent to angle $A$. Your formula requires the slopes of two lines that intersect at angle $A$.
• Nov 16th 2013, 11:11 PM
AaPa
Re: Triangle problem in coordinate geometry
thank you!!!!!!!
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