Question

how do you solve this?

Answer 1

Answer:

x=5/2 and y=3

Step-by-step 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