What is the distance fom the point 2 3 that is perpendicular to the straight line equation of 5x -y plus 7 equals 0 showing work and answer to an appropriate degree of accuracy?

Answer 1

To find the perpendicular distance from a point to a line, find the point where the line perpendicular to the original line which passes through the given point crosses the original line and the distance between those two points (using Pythagoras):

The line perpendicular to a line with gradient m has gradient m' such that mm' = -1, ie m' = -1/m

5x - y + 7 = 0

→ y = 5x + 7

→ gradient m = 5

→ gradient of perpendicular line m' = -1/5

A line through point (x0, y0) with gradient m has equation:

y - yo = m(x - x0)

→ line perpendicular to 5x - y + 7 = 0 through (2, 3) has equation:

y - 3 = -1/5 (x - 2)

→ 5y - 15 = -x + 2

→ x + 5y - 17 = 0

which crosses the original line when:

1) 5x - y + 7 = 0

2) x + 5y - 17 = 0

using (2) rearranged to give x and substituted in (1) gives:

5 (17 - 5y) - y + 7 = 0

→ 85 - 25y - y + 7 = 0

→ 26y = 92

→ y = 3 7/13

substituting back in (2) gives:

x + 5 (3 7/13) - 17 = 0

→ x = -9/13

→ lines cross at (-9/13, 3 7/13)

→ perpendicular distance from (2, 3) to the line 5x - y + 7 = 0 is the distance from (2, 3) to the point (-9/13, 3 7/13) which is:

distance = √(change_in_x^2 + change_in_y^2)

→ distance = √((2 - -9/13)^2 + (3 - 3 7/13)^2) = (√(35^2 + 7^2))/13 = (√1274)/13

I'll leave it for you to convert the exact answer (given above) to an approximate decimal of appropriate accuracy.

Additional Information:

The distance is 2.746 rounded up to three decimal places