Which equation can be used to graph the line that passes through (6, 15) and is perpendicular to the graph of y-7=1/2(x+2)

y = -2x + 27

Step-by-step explanation:

Firstly, we shall need to reform the given line equation;

y-7 = 1/2 ( x + 2)

y - 7 = x/2 + 1

y = x/2 + 1 + 8

y = x/2 + 8

Comparing this with y = mx + c

where m is the slope, then the slope of the line is 1/2

Since the line we are looking for has a slope perpendicular to this line, it means that the product of their slopes is -1

m1 * m2 = -1

1/2 * m2 = -1

m2 = -2

So we want an equation with slope -2 passing through (6,15)

we use the point slope method here;

y-y1 = m(x-x1)

y-15 = -2(x-6)

y-15 = -2x + 12

y= -2x + 12 + 15

y = -2x + 27

. + λrh() e 1 1 2 4 4 4 4 4 3 4 4 4 4 4 + th−1(− s(x)2 dx ω + λrh−1() e 2 1 2 4 4

step-by-step explanation:

-9

step-by-step explanation: