Question

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?

Answer 1

The possible points are `(-257/34, 115/17)` and `(767/34, -13/17)`

Explanation:

Given:

`"Area" = 32"units"^2`

The endpoints of side A are `(7,1)` and `(8,5)`.

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

Find the length of the base:

`"Base" = sqrt((8-7)^2+(5-1)^2)`

`"Base" = sqrt(1^2+4^2)`

`"Base" = sqrt17`

The area of a triangle is:

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

Substitute in the values of Area and Base:

`32=(1/2)(sqrt17)("Height")`

`"Height" = 64sqrt17/17`

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:

`64sqrt17/17=sqrt((x-15/2)^2+(y-3)^2)" [1]"`

We need the slope of side A:

`m_a = (5-1)/(8-7)`

`m_a = 4`

The slope of the height is perpendicular to side A:

`m_h = -1/m_a`

`m_h = -1/4`

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

`y = -1/4(x-15/2)+3" [2]"`

With equations [1] and [2] this ugly, I am going to use WolframAlpha to solve them:

The possible points are `(-257/34, 115/17)` and `(767/34, -13/17)`

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

Answer 2

`(-257/34,115/17), or, (767/34,-13/17)`

Explanation:

We solve this Problem deploying the methods of Analytic Geometry.

Supoose that the third vertex of the triangle is `P(x,y).`

Recall that, the Area of a Triangle having vertices

`(x_1,y_1), (x_2,y_2), and, (x_3,y_3)` is `1/2|D|,` where,

`D=|(x_1,y_1,1),(x_2,y_2,1),(x_3,y_3,1)|.`

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

`32=1/2|D|, where, D=|(x,y,1),(7,1,1),(8,5,1)|`

`=x(1-5)-y(7-8)+1(35-8)=-4x+y+27.`

`:. 1/2*|-4x+y+27|=32 rArr -4x+y+27=+-64.`

`:.-4x+y=64-27=37 (ast^1) or -4x+y=-91(ast^2).`
Also, Length of side B = Length of side C

`rArr (x-7)^2+(y-1)^2=(x-8)^2+(y-5)^2`

`rArr -14x+49-2y+1=-16x+64-10y+25`

`rArr2x+8y=39 rArr 4x+16y=78................(star).`

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 `(-257/34,115/17), or,`

`(767/34,-13/17)`

Enjoy Maths.!