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.
(type your answer in slope - intercept form. simplify your answer. use integers or fractions for any numbers in the expression.)
what is the equation of line m? select the correct choice below and fill in the answer box to complete your choice.
a. line m 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.)
b. line m 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 $A(54,72)$ is $m_{OA}=\frac{72 - 0}{54-0}=\frac{4}{3}$. Since line $\ell$ is perpendicular to $OA$ at $A$, the slope of $\ell$, $m_{\ell}=-\frac{3}{4}$. Using the point - slope form $y - y_1=m(x - x_1)$ with $(x_1,y_1)=(54,72)$, we have $y - 72=-\frac{3}{4}(x - 54)$, which simplifies to $y=-\frac{3}{4}x+\frac{3}{4}\times54 + 72=-\frac{3}{4}x+\frac{162}{4}+72=-\frac{3}{4}x+\frac{162 + 288}{4}=-\frac{3}{4}x+\frac{450}{4}=-\frac{3}{4}x + \frac{225}{2}$.

Step2: Find slope of OB and equation of line m

The slope of line $OB$ with $O(0,0)$ and $B(90,0)$ is $m_{OB}=\frac{0 - 0}{90-0}=0$. A line $m$ perpendicular to $OB$ at $B(90,0)$ is a vertical line. The equation of a vertical line passing through the point $(90,0)$ is $x = 90$.

Step3: Find the intersection point C

Substitute $x = 90$ into the equation of line $\ell$: $y=-\frac{3}{4}\times90+\frac{225}{2}=-\frac{270}{4}+\frac{450}{4}=\frac{-270 + 450}{4}=\frac{180}{4} = 45$. So the intersection point $C$ is $(90,45)$.

Step4: 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 $B(90,0)$: $CB=\sqrt{(90 - 90)^2+(45 - 0)^2}=\sqrt{0 + 45^2}=45$. So $CA = CB$.

Step5: Explain why C is on the bisector of ∠AOB

Since $CA=CB$ and $C$ is the intersection of the perpendiculars from $A$ to $OA$ and from $B$ to $OB$, and points that are equidistant from the two sides of an angle lie on the angle - bisector of that angle, point $C$ is on the bisector of $\angle AOB$.

Answer:

a. Equation of $\ell$: $y=-\frac{3}{4}x+\frac{225}{2}$, Equation of $m$: $x = 90$
b. $C(90,45)$
c. Shown above that $CA = CB=45$
d. Points equidistant from two sides of an angle lie on the angle - bisector, and $CA = CB$.