Find the point on the line 4x + y = 6 that is closest to the point (−4, 2).

Answer 1

find the point on the line 4x+y=6 that is closest to the point (-4,2)

y=-4x+6

that point is (12/17, 54/17) = about (.706,3.1765)

it may help to graph the lines and equation. the line 4x+y=6 and the perpendicular line through the given point which is y-2=(1/4)(x+4) or y=x/4 +3

set the two equations equal

x/4+3 = -4x+6

17x/4 = 3

x = 3(4/17) =12/17

y = -4x+6 = -4(12/17) +6= -48/17+102/17 = 54/17

the point = (12/17, 54/17)= about (.706, 3.177)

that's the algebraic method

you could also use calculus to find the shortest distance

or you could graph the circle with the given point as the center and find where it is tangent to the given line, using a graphing calculator giving (.706.3.177) if you use (x+4)^2 +(y-2)^2 =about 23.6 as the circle's equation with radius squared= about 23.6

the calculus method uses optimization

(-4,2) to the line y=-4x+6, minimize d=distance

distance from point to the line = d

with d^2= (x+4)^2+(y-2)^2

d^2= x^2 +8x+16+ (-4x+6-2)^2

= x^2 +8x +16 +(-4x+4)^2

=x^2+8x +16 +16x^2 -32x +16

= 17x^2 -24x +32=23.6 or most any constant

17x^2 -24x+8.4 = 0

x=24/34 +/-(1/34)sqr(24^2-4(17)(8.4)

= 12/17 = midpoint between the zeros = x coordinate of the vertex = minimum point of the parabola

y= -4x+6=-4(12/17)+6(17)/17=-48/17 +112/17= 54/17= y coordinate of the minimum point of the parabola

the point = (12/17, 54/17)

Answer 2

1) Determine the normal to the given line.

2) Determine equation of line having slope of normal and passing through given point

3) Locate point of intersection of normal line and given line.