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.
d. why is c on the bisector of ∠aob?
c is not in the interior of ∠aob but is equidistant from the sides of ∠aob, so c is on the bisector of ∠aob by the converse of the angle bisector theorem.
a. by the converse of the angle bisector theorem.
b. bc is perpendicular to ob, so c is on the bisector of ∠aob by the converse of the angle bisector theorem.
c. ac is perpendicular to oa, so c is on the bisector of ∠aob by the converse of the angle bisector theorem.
d. c is the mid - point of ab, so c is on the bisector of ∠aob by the converse of the angle bisector theorem.
e. c is in the interior of ∠aob and is equidistant from the sides of ∠aob, so c is on the bisector of ∠aob by the converse of the angle bisector theorem.

Explanation:

Answer:

(a) First, find the slope of \(OA\): \(m_{OA}=\frac{72 - 0}{54 - 0}=\frac{4}{3}\). The slope of line \(l\) (perpendicular to \(OA\) at \(A\)) is \(-\frac{3}{4}\). Using the point - slope form \(y - y_1=m(x - x_1)\) with \((x_1,y_1)=(54,72)\), the equation of line \(l\) is \(y - 72=-\frac{3}{4}(x - 54)\), which simplifies to \(y=-\frac{3}{4}x+\frac{3}{4}\times54 + 72=-\frac{3}{4}x+\frac{162}{4}+72=-\frac{3}{4}x+\frac{162 + 288}{4}=-\frac{3}{4}x+\frac{450}{4}=-\frac{3}{4}x + 112.5\). The slope of \(OB\) is \(m_{OB}=\frac{0 - 0}{90 - 0}=0\). The slope of line \(m\) (perpendicular to \(OB\) at \(B\)) is undefined. The equation of line \(m\) is \(x = 90\).
(b) Substitute \(x = 90\) into the equation of line \(l\): \(y=-\frac{3}{4}\times90+112.5=-67.5 + 112.5 = 45\). So the intersection point \(C\) is \((90,45)\).
(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 \(B(90,0)\): \(CB=\sqrt{(90 - 90)^2+(45 - 0)^2}=45\). So \(CA = CB\).
(d) E. C is in the interior of \(\angle AOB\) and is equidistant from the sides of \(\angle AOB\), so C is on the bisector of \(\angle AOB\) by the Converse of the Angle Bisector Theorem.