Find the distance from point (3,-5) to the line 2x+y+4=0

Answer 1

Rewriting the equation in slope-intercept form, we get Y = -2X -4. So we know there's a point on the line at (0,-4). Using the -2 slope, we go down 2 units and right 1 unit to get a point at (1,-6). Connecting the points, we see that this line is the hypoteneus of a 2-down by 1-across right triangle. We can use a proportion to find the midpoint of the hypoteneus:

1/-6 = X/-5, -6X = -5, X = 5/6

So there's a point on our line at (5/6,-5). This point is 2 and 1/6 from the problem's point of (3,-5).

Answer 2

n the case of a line in the plane given by the equation ax + by + c = 0, where a, b and c are real constants with a and b not both zero, the distance from the line to a point (x₀,y₀) is

|ax0+byo+C|/sqrt(a^2+b^2)

a=2

b=1

c=4

xo=3

y0=-5

=(6-5+4)/sqrt(4+1) 

=5/2

=2.5