Answer:
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]
