# An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from (7 ,1 ) to (8 ,5 ) and the triangle's area is 32 , what are the possible coordinates of the triangle's third corner?

May 14, 2017

The possible points are $\left(- \frac{257}{34} , \frac{115}{17}\right)$ and $\left(\frac{767}{34} , - \frac{13}{17}\right)$

#### Explanation:

Given:

${\text{Area" = 32"units}}^{2}$

The endpoints of side A are $\left(7 , 1\right)$ and $\left(8 , 5\right)$.

Let side $A =$ the base of the triangle.

Find the length of the base:

$\text{Base} = \sqrt{{\left(8 - 7\right)}^{2} + {\left(5 - 1\right)}^{2}}$

$\text{Base} = \sqrt{{1}^{2} + {4}^{2}}$

$\text{Base} = \sqrt{17}$

The area of a triangle is:

"Area"=(1/2)("Base")("Height")

Substitute in the values of Area and Base:

$32 = \left(\frac{1}{2}\right) \left(\sqrt{17}\right) \left(\text{Height}\right)$

$\text{Height} = 64 \frac{\sqrt{17}}{17}$

Compute midpoint of side A

x_"mid"= (x_"end"+x_"start")/2 = (8+7)/2 = 15/2

y_"mid"= (y_"end"+y_"start")/2 = (5+1)/2 = 3

Find the two points that are the same distance as height away from the midpoint:

$64 \frac{\sqrt{17}}{17} = \sqrt{{\left(x - \frac{15}{2}\right)}^{2} + {\left(y - 3\right)}^{2}} \text{ }$

We need the slope of side A:

${m}_{a} = \frac{5 - 1}{8 - 7}$

${m}_{a} = 4$

The slope of the height is perpendicular to side A:

${m}_{h} = - \frac{1}{m} _ a$

${m}_{h} = - \frac{1}{4}$

Use the point slope form of the equation of a line to write the equation for the height:

$y = - \frac{1}{4} \left(x - \frac{15}{2}\right) + 3 \text{ }$

With equations  and  this ugly, I am going to use WolframAlpha to solve them:

The possible points are $\left(- \frac{257}{34} , \frac{115}{17}\right)$ and $\left(\frac{767}{34} , - \frac{13}{17}\right)$

Here is an image of all of the parts of this problem that should show you that the answer is correct: May 14, 2017

$\left(- \frac{257}{34} , \frac{115}{17}\right) , \mathmr{and} , \left(\frac{767}{34} , - \frac{13}{17}\right)$

#### Explanation:

We solve this Problem deploying the methods of Analytic Geometry.

Supoose that the third vertex of the triangle is $P \left(x , y\right) .$

Recall that, the Area of a Triangle having vertices

$\left({x}_{1} , {y}_{1}\right) , \left({x}_{2} , {y}_{2}\right) , \mathmr{and} , \left({x}_{3} , {y}_{3}\right)$ is $\frac{1}{2} | D | ,$ where,

$D = | \left({x}_{1} , {y}_{1} , 1\right) , \left({x}_{2} , {y}_{2} , 1\right) , \left({x}_{3} , {y}_{3} , 1\right) | .$

Hence, in our case, since the area is $32 ,$ we have, for

$32 = \frac{1}{2} | D | , w h e r e , D = | \left(x , y , 1\right) , \left(7 , 1 , 1\right) , \left(8 , 5 , 1\right) |$

$= x \left(1 - 5\right) - y \left(7 - 8\right) + 1 \left(35 - 8\right) = - 4 x + y + 27.$

$\therefore \frac{1}{2} \cdot | - 4 x + y + 27 | = 32 \Rightarrow - 4 x + y + 27 = \pm 64.$

$\therefore - 4 x + y = 64 - 27 = 37 \left({\ast}^{1}\right) \mathmr{and} - 4 x + y = - 91 \left({\ast}^{2}\right) .$
Also, Length of side B = Length of side C

$\Rightarrow {\left(x - 7\right)}^{2} + {\left(y - 1\right)}^{2} = {\left(x - 8\right)}^{2} + {\left(y - 5\right)}^{2}$

$\Rightarrow - 14 x + 49 - 2 y + 1 = - 16 x + 64 - 10 y + 25$

$\Rightarrow 2 x + 8 y = 39 \Rightarrow 4 x + 16 y = 78. \ldots \ldots \ldots \ldots \ldots \left(\star\right) .$

Solving (ast^1) & (star), y=115/17, x=-257/34, while,

(ast^2) & (star) rArr y=-13/17, x=767/34.

Thus, the possible third vertex can be $\left(- \frac{257}{34} , \frac{115}{17}\right) , \mathmr{and} ,$

$\left(\frac{767}{34} , - \frac{13}{17}\right)$

Enjoy Maths.!