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Find the point on the line 6x+y=9 that is closest to the point (-3,1)

Minimize distance b/w (x,y) on line y=9-6x and (-3,1)

Use distance formula D=√[(x2-x1)^2+(y2-y1)^2] plug in coordinates D=√[(x+3)^2+(y-1)^2] plug in y=9-6x D=√[(x+3)^2+(8-6x)^2] multiply out and simplify D=√[37x^2-90x+73] take derivative and set equal to zero (find critical points) D/dx=.5(37x^2-90x+73)^-.5*(74x-90)=0 [74x-90]/[2√(37x^2-90x+73)]=0 74x-90=0 x=1.216

Plug into y=9-6x y=1.703

(1.216, 1.703)

To prove this is the absolute minimum distance, you must take the endpoints of the distance function into consideration. The domain is all real numbers, so we take the limit as the distance function approaches infinity and negative infinity. Both of those are equal to positive infinity, so x=1.216 is the absolute minimum, and therefore closest to the point (-3, 1)