Friday

September 4, 2015
Posted by **Robin** on Monday, July 15, 2013 at 8:47am.

- Calculus -
**MathMate**, Monday, July 15, 2013 at 9:16amThe 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 :