Question

line a has a slope of $-\frac{2}{5}$ and passes through the point $(2, 6)$.
line b has a slope of $\frac{1}{3}$ and passes through the point $(-33, -2)$.
find the point where line a intersects line b.
\\(\circ\\) a. $x = -3, y = 8$
\\(\circ\\) b. $x = -1, y = 8\frac{2}{3}$
\\(\circ\\) c. $x = 17, y = 0$
\\(\circ\\) d. $x = 8, y = -3$

Explanation:

Step1: Find equation of Line A

Using point - slope form \(y - y_1=m(x - x_1)\), where \(m =-\frac{2}{5}\), \(x_1 = 2\), \(y_1 = 6\).
\(y-6=-\frac{2}{5}(x - 2)\)
\(y-6=-\frac{2}{5}x+\frac{4}{5}\)
\(y=-\frac{2}{5}x+\frac{4}{5}+6\)
\(y=-\frac{2}{5}x+\frac{4 + 30}{5}\)
\(y=-\frac{2}{5}x+\frac{34}{5}\)

Step2: Find equation of Line B

Using point - slope form \(y - y_1=m(x - x_1)\), where \(m=\frac{1}{3}\), \(x_1=- 33\), \(y_1=-2\).
\(y + 2=\frac{1}{3}(x + 33)\)
\(y+2=\frac{1}{3}x + 11\)
\(y=\frac{1}{3}x+11 - 2\)
\(y=\frac{1}{3}x + 9\)

Step3: Set equations equal to find intersection

Set \(-\frac{2}{5}x+\frac{34}{5}=\frac{1}{3}x + 9\)
Multiply through by 15 to eliminate denominators:
\(15\times(-\frac{2}{5}x)+15\times\frac{34}{5}=15\times\frac{1}{3}x+15\times9\)
\(- 6x + 102 = 5x+135\)
\(-6x-5x=135 - 102\)
\(-11x = 33\)
\(x=-3\)

Step4: Find y - value

Substitute \(x = - 3\) into \(y=\frac{1}{3}x + 9\)
\(y=\frac{1}{3}(-3)+9=-1 + 9 = 8\)

Answer:

A. \(x=-3,y = 8\)