What is an equation of the line through point (2,1) and perpendicular to the line (-4,1) and (3, -2)?

Gradient of the line with points (-4,1)&(3,-2) is

(-2-1)÷(3-(-4))=-3/7

Then the equation of that^ line is

Y-1=-3/7(x-(-4))--->y=-3/7x-5/7

Since this^ line is perpendicular to the line you are looking for.the gradient of the line you are looking for is -1÷(-3/7)=7/3

The equation of the line you are looking for is

Y-1=7/3(x-2)--->y=7/3x-11/3

[Cause the line you are looking for passes through the point (2,1) and you found the gradient of this line]

answer: group b.

step-by-step explanation:

multiple all that

step-by-step explanation:

27×17