Find the distance of a line

Question

Line ℓ has equation y=x–5

.

 Find the distance between ℓ and the point  T(-7,4)

.

Answer 1

The distance from a point, (x₀,y₀) to a line is the length of a line segment from the point to the line that is perpendicular to that line. If the equation OF THAT LINE IS Ax + By + C = 0, then the distance d is given by the formula:

 

d = |Ax₀ + By₀ + C| / sqrt(A² + B²)          eq(1)

In this case, the equation of the line y=x-5 is the equivalent of x – y – 5 = 0, so:

A= 1

B = -1

C = -5, and:

x₀ = -7

y₀ = 4

Plugging these values into eq(1):

d = |1*(-7) + (-1)*4 - 5| / sqrt(1² + (-1)²)

  = |-7 - 4 – 5| / sqrt(2)

  = 15/sqrt(2)

  = 10.6061 (approximated)

Answer 2

The slope of the line y = x-5 is 1.

Equation of line perpendicular to y = x-5 and passing through (-7,4) is:

y - 4 = -1(x+7)

y = -x-3

The line y = x-5 intersects the line y = -x-3 when x - 5 = -x -3

2x = 2

x = 1

The lines intersect at the point P = (1, -4).

Distance between the line y = x-5 and the point T is the same as the distance, D, between P and T.

D = √[(1-(-7))² + (-4-4)²] = √128 = 8√2