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.
b. the intersection c of lines l and m is (90,45). (type an ordered pair.)
c. find ca.
ca = 45 (type an exact answer, using radicals as needed.)
find cb.
cb = (type an exact answer, using radicals as needed.)

Explanation:

Answer:

(a) First, find the slope of \(OA\): \(m_{OA}=\frac{72 - 0}{54 - 0}=\frac{4}{3}\). The slope of line \(\ell\) (perpendicular to \(OA\) at \(A\)) is \(m_{\ell}=-\frac{3}{4}\). Using the point - slope form \(y - y_1=m(x - x_1)\) with \((x_1,y_1)=(54,72)\), the equation of \(\ell\) is \(y - 72=-\frac{3}{4}(x - 54)\), which simplifies to \(y=-\frac{3}{4}x+\frac{3\times54}{4}+72=-\frac{3}{4}x+\frac{162 + 288}{4}=-\frac{3}{4}x+\frac{450}{4}=-\frac{3}{4}x + 112.5\). The slope of \(OB\) is \(m_{OB}=0\), so the slope of line \(m\) (perpendicular to \(OB\) at \(B\)) is undefined. The equation of \(m\) is \(x = 90\).
(b) Substitute \(x = 90\) into the equation of \(\ell\): \(y=-\frac{3}{4}\times90+112.5=-67.5 + 112.5 = 45\). So \(C=(90,45)\)
(c) Using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), 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 \(B(90,0)\): \(CB=\sqrt{(90 - 90)^2+(45 - 0)^2}=45\)
(d) Since \(CA = CB\), point \(C\) is equidistant from the two sides of \(\angle AOB\). By the converse of the angle - bisector theorem, which states that if a point is equidistant from the two sides of an angle, then it lies on the angle - bisector of the angle, \(C\) is on the bisector of \(\angle AOB\)

So:
(a) Equation of \(\ell\): \(y=-\frac{3}{4}x + 112.5\), Equation of \(m\): \(x = 90\)
(b) \((90,45)\)
(c) \(CA = 45\), \(CB = 45\)
(d) Point \(C\) is equidistant from the two sides of \(\angle AOB\), so by the converse of the angle - bisector theorem, \(C\) is on the bisector of \(\angle AOB\)