Posted by Robin on Monday, July 15, 2013 at 8:47am.
The idea is to find the family of lines perpendicular to L:7x+5y-2=0.
Out of that, we find the particular line L1 passing through P(-6,-2).
The intersection of L and L1 gives the required point (closest to P).
Family of lines perpendicular to L passing through P0(x0,y0):
L : 7x+5y-2=0
L1 : 5(x-x0)-7(y-y0)=0
in the above, we switch the coefficients of x and y, and change the sign of one of them. To make the line pass through P0, we subtract the coordinates of P0 from x and y respectively.
For P0(-6,-2),
L1 : 5(x-(-6))-7(y-(-2))=0
=> 5x-7y+16=0
Finally, the required point is the intersection point of L and L1, namely, the solution of the system of equations
7x+5y-2=0 and
5x-7y+16=0
Solving by elimination or by Cramer's rule, we obtain: (-33/37, 61/37)
Substitute values to verify that the solution is correct:
7x+5y=7(-33/37)+5(61/37)=2
5x-7y=5(-33/37)-7(61/37)=-16
ok
L1 :