Question

Two corners of an isosceles triangle are at `(1 ,7 )` and `(5 ,3 )`. If the triangle's area is `6`, what are the lengths of the triangle's sides?

Answer 1

Let the coordinates of the third corner of the isosceles triangle be `(x,y)`. This point is equidistant from other two corners.

So

`(x-1)^2+(y-7)^2=(x-5)^2+(y-3)^2`

`=>x^2-2x+1+y^2-14y+49=x^2-10x+25+y^2-6y+9`

`=>8x-8y=-16`

`=>x-y=-2`

`=>y=x+2`

Now the perpendicular drawn from `(x,y)` on the line segment joining two given corners of triangle will bisect the side and the coordinates of this mid point will be `(3,5)`.

So height of the triangle

`H=sqrt((x-3)^2+(y-5)^2)`

And base of the triangle

`B=sqrt((1-5)^2+(7-3)^2)=4sqrt2`

Area of the triangle

`1/2xxBxxH=6`

`=>H=12/B=12/(4sqrt2)`

`=>H^2=9/2`

`=>(x-3)^2+(y-5)^2=9/2`

`=>(x-3)^2+(x+2-5)^2=9/2`

`=>2(x-3)^2=9/2`

`=>(x-3)^2=9/4`

`=>x=3/2+3=9/2=4.5`

So `y=x+2=4.5+2=6.5`

Hence length of each equal sides

`=sqrt((5-4.5)^2+(3-6.5)^2)`

`=sqrt(0.25+12.25)=sqrt12.5=2.5sqrt2`

Hence lengths of three sides are `2.5sqrt2,2.5sqrt2,4sqrt2`