Answer:
The distance is [tex]\sqrt{10}[/tex]
Step-by-step explanation:
First we need to find the slope of the line and the the its equation, for the slope we use the formula:
[tex]m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{-4-1}{9-(-6)}\\ m=\frac{-5}{9+6}\\ m=\frac{-5}{15}\\m=- \frac{1}{3}[/tex]
Using the line equation we have that:
[tex]y-y_1=m(x-x_1)\\y-1=- \frac{1}{3} (x-(-6))\\y=- \frac{1}{3} (x+6)+1\\y=- \frac{1}{3} x-2+1\\ y=-\frac{1}{3} -1x [/tex] (1)
Converting the line equation (1) into its general form
[tex]y=- \frac{1}{3} x-1\\3y=-x-3 \\x+3y+3=0[/tex]
Finally we use the distance between a line and point formula given by
[tex]d=\frac{Ax+By+C}{ \sqrt{ A^{2} + B^{2} } }[/tex]
where A, B and C are the coefficients of the line and x and y are the coordinates of the point P
[tex]d=\frac{1(4)+3(1)+3}{ \sqrt{ 1^{2}+3^{2} } } \\d=\frac{4+3+3}{ \sqrt{1+9} }\\d=\frac{10}{ \sqrt{10} }\\d=\frac{10 \sqrt{10} }{10}\\d=\sqrt{10}[/tex]
