Question

use points a(120,64), o(0,0), and b(136,0).
a. write equations of lines l and m such that l⊥oa at a and m⊥ob at b.
b. find the intersection c of lines l and m.
c. show that ca = cb.
d. explain why c is on the bisector of ∠aob.
a. what is the equation of line l? select the correct choice below and fill in the answer box to complete your choice.
a. line l is not vertical. its equation is y = -15/8 + 289. (type your answer in slope - intercept form. simplify your answer. use integers or fractions for any numbers in the expression.)
b. line l is vertical. its equation is x = (type an integer or a simplified fraction.)

Explanation:

Step1: Find slope of OA

The slope of line OA with $O(0,0)$ and $A(120,64)$ is $m_{OA}=\frac{y_A - y_O}{x_A - x_O}=\frac{64 - 0}{120 - 0}=\frac{8}{15}$. Since line $\ell\perp OA$ at $A$, the slope of line $\ell$ is the negative - reciprocal of the slope of OA. So $m_{\ell}=-\frac{15}{8}$.

Step2: Use point - slope form to find equation of line $\ell$

The point - slope form of a line is $y - y_1=m(x - x_1)$. Using point $A(120,64)$ and $m =-\frac{15}{8}$, we have $y - 64=-\frac{15}{8}(x - 120)$. Expanding gives $y-64=-\frac{15}{8}x+\frac{15\times120}{8}=-\frac{15}{8}x + 225$. Then $y=-\frac{15}{8}x+289$. So line $\ell$ is not vertical.

Answer:

A. Line $\ell$ is not vertical. Its equation is $y =-\frac{15}{8}x + 289$