Question

a finish line banner is three times as long as the distance from a cone at point j to the race path defined by y = -2/3 x. find the price of the banner when it costs $3 per linear foot. price: about $

Explanation:

Step1: Find the distance from \( J(6,4) \) to the line \( y =-\frac{2}{3}x \)

The formula for the distance \( d \) from a point \((x_0,y_0)\) to a line \(Ax + By+C = 0\) is \(d=\frac{\vert Ax_0+By_0 + C\vert}{\sqrt{A^{2}+B^{2}}}\).
Rewrite \(y=-\frac{2}{3}x\) as \(2x+3y = 0\). Here \(x_0 = 6\), \(y_0=4\), \(A = 2\), \(B = 3\), \(C = 0\).

$$d=\frac{\vert2\times6 + 3\times4+0\vert}{\sqrt{2^{2}+3^{2}}}=\frac{\vert12 + 12\vert}{\sqrt{4 + 9}}=\frac{24}{\sqrt{13}}$$

Step2: Calculate the length of the banner

Since the banner is three - times as long as the distance from the point to the line. Let \(L\) be the length of the banner.
\(L = 3\times\frac{24}{\sqrt{13}}=\frac{72}{\sqrt{13}}\)

Step3: Calculate the price of the banner

The cost is \( \$3\) per linear foot. Let \(P\) be the price.
\(P=3\times\frac{72}{\sqrt{13}}=\frac{216}{\sqrt{13}}\approx\frac{216}{3.606}\approx60\)

Answer:

\(60\)