We are given two points of a line
(-2, -2) and (1,4)
Equation of line:
Firstly , we will find slope
(-2, -2) and (1,4)
[tex] m=\frac{y_2-y_1}{x_2-x_1} [/tex]
[tex] m=\frac{4+2}{1+2} [/tex]
[tex] m=2 [/tex]
now, we can find equation of line
[tex] y-y_1=m(x-x_1) [/tex]
now, we can plug values
and we get
[tex] y+2=2(x+2) [/tex]
[tex] 2x-y+2=0 [/tex]
Distance:
now, we can use distance formula
[tex] d=\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}} [/tex]
firstly , we will find a and b
a=2 and b=-1
point is (6,-1)
so, xo=6 and yo=-1
we can plug values
and we get
[tex] d=\frac{|2*6-1*-1+2|}{\sqrt{(2)^2+(-1)^2}} [/tex]
[tex] d=\frac{15}{\sqrt{5}} [/tex]
[tex] d=6.70820 [/tex]...........Answer
