Answer:
The distance from p to l is [tex]\sqrt{10}\ units[/tex]
Step-by-step explanation:
we know that
The distance between point p from line l is equal to the perpendicular segment from line l to point p
step 1
Find the slope of line l
we have the points
(1,5) and (4, -4)
The formula to calculate the slope between two points is equal to
[tex]m=\frac{y2-y1}{x2-x1}[/tex]
substitute the values
[tex]m=\frac{-4-5}{4-1}[/tex]
[tex]m=\frac{-9}{3}[/tex]
[tex]m=-3[/tex]
step 2
Find the equation of the line l
The equation in point slope form is equal to
[tex]y-y1=m(x-x1)[/tex]
we have
[tex]m=-3[/tex]
[tex]point\ (1,5)[/tex]
substitute
[tex]y-5=-3(x-1)[/tex] -----> equation A
step 3
Find the slope of the line perpendicular to the line l
Remember that
If two lines are perpendicular, then their slopes are opposite reciprocal (The product of their slopes is equal to -1)
[tex]m_1*m_2=-1[/tex]
we have
[tex]m_1=-3[/tex] ---> slope of line l
therefore
[tex]m_2=\frac{1}{3}[/tex] ----> slope of the line perpendicular to line l
step 4
Find the equation of the line perpendicular to line l that passes through the point p
The equation in point slope form is equal to
[tex]y-y1=m(x-x1)[/tex]
we have
[tex]m=\frac{1}{3}[/tex]
[tex]point\ p(-1,1)[/tex]
substitute
[tex]y-1=\frac{1}{3}(x+1)[/tex] -----> equation B
step 5
Solve the system of equations
[tex]y-5=-3(x-1)[/tex] -----> equation A
[tex]y-1=\frac{1}{3}(x+1)[/tex] -----> equation B
Solve the system by graphing
The solution of the system is the intersection point both graphs
The solution is the point q(2,2)
see the attached figure
step 6
we know that
The distance between the point p and the line l is equal to the distance between the point p and the point q
the formula to calculate the distance between two points is equal to
[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]
we have the points
p(-1,1) and q(2,2)
substitute the values
[tex]d_p_q=\sqrt{(2-1)^{2}+(2+1)^{2}}[/tex]
[tex]d_p_q=\sqrt{(1)^{2}+(3)^{2}}[/tex]
[tex]d_p_q=\sqrt{10}\ units[/tex]
