Question

i need the equation of the line that satisfies the following conditions: it passes through the point of intersection of the curves y=x^2-8x+8 and 2+x=y+2 and it is also perpendicular to the equation 2y-3+6x=0

Answer 1

I'm going to try and help for this one give me a sec

Answer 2

FOR NOW YOUR SLOPE IS THE NEGATIVE RECIPROCAL OF THE EQUATION 2Y-3+6X=0
WHICH IS 1/3 BECAUSE SLOPE OF THE EQUATION ABOVE IS -3 WHEN RE-WRITTEN.

Answer 3

so if i find the intersection by substituting y=x^2-8x+8 to 2 +x=y+2? and then use those points to find my equation?

Answer 4

RIGHT,
THE OTHER TWO EQUATIONS INTERSECT AT (1,1) AKA (X1,Y1) AND THE EQUATION FOR A LINE THAT INTERSECTS A POINT IS Y-Y1=M(X-X1)

Answer 5

but what about the other point

Answer 6

ive got to intersections

Answer 7

RIGHT! (8,8) YOU CAN USE EITHER POINT.. POINT INTERCEPT FORM OF AN EQUATION ONLY REQUIRES ONE SET OF POINTS AND THE SLOPE M=1/3.. SO HOPE I HELPED A BIT

Answer 8

I MEAN POINT SLOPE HAHA

Answer 9

ok ok
thanks man

Answer 10

i got it from here

Answer 11

RIGHT ON! GOOD LUCK

Answer 12

y=x/3+2/3 or y=x/3+16/3