The distance between line l and point p is 5 units.
Given:
Line l contains points \((0,-2)\) and \((6,6)\).
Point P has the coordinates \((-1,5)\).
To find:
The distance between line l and point P.
Explanation:
The equation of the line l is:
\(y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)\)
\(y-(-2)=\dfrac{6-(-2)}{6-0}(x-0)\)
\(y+2=\dfrac{8}{6}x\)
Multiply both sides by \(6\).
\(6(y+2)=8x\)
\(6y+12=8x\)
\(0=8x-6y-12\)
The distance between the line \(ax+by+c=0\) and the point \((x_0,y_0)\) is:
\(D=\dfrac{ax_0+by_0+c}{\sqrt{a^2+b^2}}\)
The distance between the line \(8x-6y-12=0\) and the point \((-1,5)\).
\(D=\dfrac{8(-1)-6(5)-12}{\sqrt{(8)^2+(6)^2}}\)
\(D=\dfrac{-8-30-12}{\sqrt{64+36}}\)
\(D=\dfrac{50}{\sqrt{100}}\)
\(D=\dfrac{50}{10}\)
\(D=5\)
Therefore, the distance between line l and point p is 5 units.
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