Find the distance from the points (6,10) to the line y=-x

Question

Find the distance from the points (6,10) to the line y=-x

Please be specific and answer the question step by step. Please make sure the steps are specific and please give me an answer after the steps

Answer 1

Find equation of line perpendicular to y = -x

slope of that line is -1, so slope of perpendicular line is 1.

equation of line is y = mx + b

substitute, 6 for x, 10 for y and 1 for m to find b:

10 = (1) 6 + b

b = 4

so equation of perpendicular line is y = x + 4

equate equations of line to find intersecting x:

x + 4 = -x

x = -2

using equation y = -x, then intersecting y is 2, so point where two lines intersect is (-2,2)

find distance between (6,10) and (-2,2):

sqrt((6--2)^2 + (10-2)^2) = sqrt( 128) = 11.31

Answer 2

If you draw a dashed-imaginary line perpendicular to the y = -x from the point(6, 10), the imaginary line has to have a slope 1 because perpendicular lines' slopes yield -1 when multiplied.

Perpendicularity : m₁ x m₂ = -1

So, this imaginary new line is in y = mx + b form where m = 1, let's plug (6,10) into it because it passes from (6,10)

10 = 1 x 6 + b ==> b = 10 - 6 = 4

Boom, we get the equation of this line. y = x + 4 ( satisfies 6,10 and slope 1 )

Also, these two lines intersect and let's find that point solving them simultaneously

y = -x

y = x + 6

x + 6 = -x ==> x = -3, then y = 3

Now, we will find the distance between (6,10) and (-3,3) as the distance is the shortest perpendicular path between the two. Just apply Pythagorean to find the distance. D² = (y₂ - y₁)² + (x₂ - x₁)²

(10-3)^2 + (6 - -3)^2 = 49 + 81 = 130. Then D = √130=∼11.4

Please double check