Calculus
posted by Robin .
Find the point on the line 7x+5y−2=0 which is closest to the point (−6,−2).

The idea is to find the family of lines perpendicular to L:7x+5y2=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+5y2=0
L1 : 5(xx0)7(yy0)=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
=> 5x7y+16=0
Finally, the required point is the intersection point of L and L1, namely, the solution of the system of equations
7x+5y2=0 and
5x7y+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
5x7y=5(33/37)7(61/37)=16
ok
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