Calculus Question: Find the point on the line 2x+4y+3=0 which is closest to the point (-1,-1)?

The shortest straight line from the point (-1, -1) to the line 2x + 4y + 3 = 0 meets it at right-angles.

If 2 lines are perpendicular then the product of their slopes is - 1.

2x + 4y + 3 = 0 can be rearranged as

4y = - 2x - 3

y = (-1/2)x - 3/4

So the slope of the line 2x + 4y + 3 = 0 is -1/2

and the slope of the shortest line is -1/(-1/2) = 2.

The equation of the shortest line is

y - (-1) = 2[x - (-1)]

y + 1 = 2(x + 1) = 2x + 2

y = 2x + 1

The closest point to (-1, -1) on the line 2x + 4y + 3 = 0 is found by solving

2x + 4y + 3 = 0 and

y = 2x + 1

simultaneously.

From the 2nd equation, we get 2x = y - 1

Substituting this into the first equation, we get

y - 1 + 4y + 3 = 0

5y = - 2

y = - 2/5

Rearranging the 2nd equation, we get x = (y - 1)/2 = - 7/10

The closest point is (-7/10, -2/5) or (-0.7, -0.4).

2x+4y+3 = 0

y = -½x - ¾

slope = -½

slope of any perpendicular to line = 2

line of slope 2 that passes through (-1,-1):

y+1 = 2(x+1)

y = 2x + 1

intersection of lines:

2x + 1 = -½x - ¾

2.5x = -1.75

x = -0.7

(-1,-1) is closest to (-0.7, -0.4)

hints: closest distance is always perpendicular to the line , so , u find the equation of the perpendicular , then find the intersection of the 2 lines .

another hint. in such questions, always draw to imagine the case . it helps alot :)

good luck