Find the point on the line 2x+6y+3=0...

Question

Find the point on the line 2x+6y+3=0 which is closest to the point (−4,4).

please include steps in the answer! I have used the distance formula, calculated the derivative, and set it equal to zero, but I am getting "no solution."

any help is appreciated!

Answer 1

answer is (-4.95, 1.15)

2x+6y+3=0 is the given line

rewrite in slope intercept form

6y=-2x -3

y=-2x/6-3/6

y=-x/3 -1/2 with slope =-1/3

a perpendicular line has negative inverse slope = +3/1= 3

take -1/3, flip it upside down and change the sign

through (-4,4) the perpendicular line is

y-4= 3(x- -4) in point slope form

y=3(x+4)+4

y=3x+16

it may help to graph the line and plot the points

if only a rough sketch

they intersect when

y=y

3x+16=-x/3-1/2

3x+ x/3 = -1/2 -16

10x/3 = -33/2

x = (-33/2)(3/10) =-99/20= -4.95= about -5

y = 3(-99/20)+16

y= -297/20 +16

y =-297/20 + 320/20

y = 23/20= 1.15= about 1

intersection point is (-99/20, 23/20)= (-4.95, 1.15) = about (-5,1)

which is the closest point on the line to the point (-4,4)

find the distance from the intersection point to (-4,4)= (-80/20, 80/20)

d^2 = (19/20)^2 + (57/20)^2

d = (1/20)sqr(19^2 +57^2)

= (1/20)sqr(361+3249)

= (1/20)sqr3610

= about 3.004163777

= about 3

(-5,1) to (-4,4) has d=

sqr(1^2+3^2) = sqr10

= about 3.16

3 is the right range for the answer, given the rounding to get (-5,1)

the shortest distance from a point to a line is

via a perpendicular line, find where the 2 lines

intersect and calculate the distance between

those 2 points

from (-4.95, 1.15) to (-4,4)

d^2 = .95^2 + 2.85^2

= .9025+8.1225

=9.025

d = sqr9.025

= about 3.00416= shortest distance from the line to the point (-4,4)

look at the rough sketch of the lines and points to feel comfortable with the solution

or you could use calculus, derivatives, an optimization minimization problem

and get totally lost