Question

Two corners of an isosceles triangle are at `(7 ,4 )` and `(3 ,1 )`. If the triangle's area is `64`, what are the lengths of the triangle's sides?

Answer 1

the lengths are `5` and `1/50sqrt(1654025)=25.7218`
and `1/50sqrt(1654025)=25.7218`

Explanation:

Let `P_1(3, 1), P_2(7, 4), P_3(x, y)`

Use the formula for the area of a polygon

`Area=1/2((x_1,x_2,x_3,x_1),(y_1,y_2,y_3,y_1))`

`Area=1/2(x_1y_2+x_2y_3+x_3y_1-x_2y_1-x_3y_2-x_1y_3)`

`64=1/2((3,7,x,3),(1,4,y,1))`

`128=12+7y+x-7-4x-3y`

`3x-4y=-123" "`first equation

We need a second equation which is the equation of the perpendicular bisector of the segment connecting `P_1(3, 1),and P_2(7, 4)`

the slope `=(y_2-y_1)/(x_2-x_1)=(4-1)/(7-3)=3/4`

for the perpendicular bisector equation, we need slope`=-4/3` and the midpoint `M(x_m, y_m)` of `P_1` and `P_2`

`x_m=(x_2+x_1)/2=(7+3)/2=5`
`y_m=(y_2+y_1)/2=(4+1)/2=5/2`

Perpendicular bisector equation
`y-y_m=-4/3(x-x_m)`
`y-5/2=-4/3(x-5)`
`6y-15=-8x+40`
`8x+6y=55" "`second equation

Simultaneous solution using first and second equations
`3x-4y=-123" "`
`8x+6y=55" "`

`x=-259/25` and `y=1149/50`
and `P_3(-259/25, 1149/50)`

We can now compute for the other sides of the triangle using distance formula for `P_1` to `P_3`

`d=sqrt((x_1-x_3)^2+(y_1-y_3)^2)`

`d=sqrt((3--259/25)^2+(1-1149/50)^2)`

`d=1/50sqrt(1654025)`
`d=25.7218`

We can now compute for the other sides of the triangle using distance formula for `P_2` to `P_3`

`d=sqrt((x_2-x_3)^2+(y_2-y_3)^2)`

`d=sqrt((7--259/25)^2+(4-1149/50)^2)`

`d=1/50sqrt(1654025)`
`d=25.7218`

God bless...I hope the explanation is useful.