Question

How do you solve this?

Answer 1

x=5/2 and y=3

Explanation:

since AC is a perpendicular bisector, BC=CD

`6x-6=3y` so
`6x=3y+6`

if AC = a

using pythagorean theorem:

`(6x-6)^2 + a^2 = (6x+1)^2`

`(3y)^2 + a^2 = (5y+1)^2`

simplifies to:

`36x^2-72x+36+a^2=36x^2+12x+1`

`9y^2+a^2=25y^2+10y+1`

further simplifies to:

`84x-35=a^2`

`16y^2+10y+1=(8y+1)(2y+1)=a^2`

so we now have

`84x-35=(8y+1)(2y+1)`

plugging the first equation in we get

`42y+84-35=(8y+1)(2y+1)=16y^2+10y+1`

`42y+49=16y^2+10y+1`

`16y^2-32y-48=0`

`y^2-2y-3=0`

`(y-3)(y+1)=0` so
`y=-1, y=3`

y=-1 is impossible, so y=3

therefore x=15/6=5/2