Question

find the distance between point p and line (ell). line (ell) contains points ((11, -1)) and ((-3, -11)). point p has coordinates ((-1, 1)). a) (sqrt{74}) or about 8.60 units b) (sqrt{76}) or about 8.72 units c) (sqrt{47}) or about 6.86 units d) 8 units

Explanation:

Step1: Find the equation of line \( \ell \)

The slope \( m \) of the line passing through \((x_1,y_1)=(11, - 1)\) and \((x_2,y_2)=(-3,-11)\) is \(m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{-11+1}{-3 - 11}=\frac{-10}{-14}=\frac{5}{7}\).
Using the point - slope form \(y - y_1=m(x - x_1)\) with \((x_1,y_1)=(11,-1)\), we get \(y+1=\frac{5}{7}(x - 11)\).
Multiply through by \(7\) to get \(7y+7 = 5x-55\), or \(5x-7y-62 = 0\).

Step2: Use the distance formula

The distance \(d\) from a point \(P(x_0,y_0)=(-1,1)\) to a line \(Ax+By+C = 0\) (here \(A = 5\), \(B=-7\), \(C=-62\)) is given by \(d=\frac{\vert Ax_0+By_0+C\vert}{\sqrt{A^{2}+B^{2}}}\).
Substitute \(x_0=-1\), \(y_0 = 1\) into the formula: \(d=\frac{\vert5\times(-1)-7\times1-62\vert}{\sqrt{5^{2}+(-7)^{2}}}=\frac{\vert-5 - 7-62\vert}{\sqrt{25 + 49}}=\frac{\vert-74\vert}{\sqrt{74}}=\sqrt{74}\approx8.60\).

Answer:

A. \(\sqrt{74}\) or about \(8.60\) units