# Help solving? What are the possible values of x and y? Line AB is perpendicular to Line CD. A(1,3). B(4, -2). C(6,1). D(x, y). Answer choices below.A)x=1, y=-2 B)x=3, y=6 C)x=3, y=-4 D)z=-2, y=-4

hala718 | Certified Educator

Since AB is perpendicular to AC

Then the product of the slopes = -1

Let m1 be the slope for AB:

==> m1= (-2-3)/(4-1) = -5/3

Let m2 be the slope for CD

==> m1*m2 = -1

==> -5/3 *m2 = -1

==: m2 = 3/5

but m2 = (y-1)/(x-6) = 3/5

==> y-1 = (3/5)(x-6)

==> y= (3/5)(x-6) + 1

=> y= (3/5)x - 18/5 + 5/5

==> y= (3/5)x - 13/5

Therfore , we need x, y alues that verifies the equation,

Let us substitute the choices:

A( 1, -2)

==> y= (3/5)*1 - 13/5

= 3/5 - 13/5 = -10/5 = -2

The point (1, -2) verifies the equation.

Then D = (1, -2)

neela | Student

AB is perpendicular to CD

A(1,3). B(4, -2). C(6,1). D(x, y).

Sope od AB = (By -Ay)/(Bx-Ax) = (-2-3)/((4-1) =  -5/3

Slope of CD = (Dy-Cy)/(Dx -Cy) = (y-1)/(x-6).

For AB and CD to be perpendicular, the product of their slope should be equal to -1.

So (-5/3)(y-1)/(x-6) = -1

-5(y-1) = 3(x-6 )(-1)

-5y +5 = 18-3x.

3x-5y +5-18 = 0.

3x-5y- 13 = 0.

Therefore  any point onthe line 3x-5y - 13 = 0 which passes through (6,1)

3x-5y - 13 = 0.