Find the distance from P(14, –6) to line 𝓁 that contains points (–2, 4) and (1, –9). Express your answer in simplified radical form, if necessary.

Answer 1

P(14, –6) to line 𝓁 that contains points (–2, 4) and (1, –9).

line has equation -13/3=(y-4)/(x+2)...y=(-13/3)x-(26/3)+12/3=(-13/3)x-14/3

when x=14,,,y=(14/3)(-13-1)=-14²/3

from similar triangles, given (x₀,y₀) and line y=mx+b, distance from line to point is

d=|mx₀+b-y₀|/√(1+m²),,,,,√(1+m²) is hypotenuse with leg=1,,,|mx₀+b-y₀| is hypotenuse with leg=d

so d=|-(14²/3)+6|/√(1+(-13/3)²)

corrected arithmetic

=(178/3)/((√178)/3)=178/√178=√178

Answer 2

The following steps can provide you with the answer

1.

Determine the equation of the line, l, passing through (-2, 4) and (1, -9)

2.

Determine the slope of the equation/line, l.

3.

Use the negative inverse of the slope from 2 and the point (14, -6 ) to determine the perpendicular to line, l

4.

Determine the point of intersection of the line perpendicular (3) and the line l.

5

Use the distance formula to determine the distance from the point of intersection to P.