Given p(12,-2), q(5,-10), r(-4,10) and s(4,y), find the value of y so that line pq is perpendicular to line rs

the quotient is 4x^2 - 27x + 167 with a remainder of -996

step-by-step explanation:

in synthetic division, use -6 as the divisor and use the coefficients {4, -3, 5, 6}:

-6     )   4     -3     5     6

                -24   162 -1002

          4   -27   167   -996

thus, the quotient is 4x^2 - 27x + 167 with a remainder of -996.

note:   your "4x°-3x+5x+6" is ambiguous / incomplete.   i had to assume that you meant '4x^3 - 3x^2 + 5x + 6."

the equation of the parallel line would be y = 2x + 9

step-by-step explanation:

to start we need to find the slope of the original line. we can do that by looking at the coefficient of x, which is 2

since the slope of the original line is 2 and parallel lines have the same slope, we know the new line will also have 2 slope. now we can use that along with the point given in point-slope form to get the find equation.

y - y1 = m(x - x1)

y - 5 = 2(x + 2)

y - 5 = 2x + 4

y = 2x + 9

-5, your answer was correct