Question

use points a(54,72), o(0,0), and b(90,0).
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 vertical. its equation is x =
(type an integer or a simplified fraction.)
○ b. line l is not vertical. its equation is y =
(type your answer in slope - intercept form. simplify your answer. use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Find slope of OA

The slope of line OA with $O(0,0)$ and $A(54,72)$ is $m_{OA}=\frac{y_A - y_O}{x_A - x_O}=\frac{72 - 0}{54 - 0}=\frac{4}{3}$.

Step2: Find slope of line l

Since line $l\perp OA$, the slope of line $l$, $m_l$ is the negative - reciprocal of $m_{OA}$. So $m_l=-\frac{3}{4}$.

Step3: Write equation of line l using point - slope form

Using the point - slope form $y - y_1=m(x - x_1)$ with point $A(54,72)$ and $m =-\frac{3}{4}$, we have $y - 72=-\frac{3}{4}(x - 54)$.
Expand: $y-72=-\frac{3}{4}x+\frac{162}{4}$.
$y=-\frac{3}{4}x+\frac{162}{4}+72=-\frac{3}{4}x+\frac{162 + 288}{4}=-\frac{3}{4}x+\frac{450}{4}=-\frac{3}{4}x + \frac{225}{2}$.

The slope of line OB with $O(0,0)$ and $B(90,0)$ is $m_{OB}=\frac{0 - 0}{90 - 0}=0$.
Since line $m\perp OB$, line $m$ is a vertical line. Its equation is $x = 90$.

Step4: Find the intersection point C

Set the equations of $l$ and $m$ equal. Substitute $x = 90$ into $y=-\frac{3}{4}x+\frac{225}{2}$.
$y=-\frac{3}{4}\times90+\frac{225}{2}=-\frac{270}{4}+\frac{450}{4}=\frac{-270 + 450}{4}=\frac{180}{4} = 45$. So $C(90,45)$.

Step5: Calculate CA and CB

The distance formula is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.
For $CA$ with $C(90,45)$ and $A(54,72)$:
$CA=\sqrt{(90 - 54)^2+(45 - 72)^2}=\sqrt{36^2+(-27)^2}=\sqrt{1296 + 729}=\sqrt{2025}=45$.
For $CB$ with $C(90,45)$ and $B(90,0)$:
$CB=\sqrt{(90 - 90)^2+(45 - 0)^2}=\sqrt{0 + 2025}=45$. So $CA = CB$.

Step6: Explain why C is on the bisector of $\angle AOB$

Points on the angle - bisector of an angle are equidistant from the sides of the angle. Since $CA$ is the distance from $C$ to side $OA$ and $CB$ is the distance from $C$ to side $OB$, and $CA = CB$, point $C$ is on the bisector of $\angle AOB$.

Answer:

a. B. Line $l$ is not vertical. Its equation is $y=-\frac{3}{4}x+\frac{225}{2}$
b. $C(90,45)$
c. Shown above.
d. Points on the angle - bisector are equidistant from the sides of the angle and $CA = CB$.