Exercise 28 Page 418

Angles $\angle 1$ and $\angle 2,$ together with the right angle between lines ${\ell }_1$ and ${\ell }_2,$ form a straight angle, so their measures add up to $180.$ Let's now look at triangle $△OBD.$ Two of the angles of this triangle are $\angle 2$ and $\angle 3.$ According to the Triangle Angle-Sum Theorem, the interior angle measures add up to $180.$ We now have two angle sums that both give $180.$ Let's set them equal to each other and simplify the equation. This means that $\angle 1$ and $\angle 3$ are congruent. These are angles of triangles $△OBD$ and $△OAC.$ Note that these triangles are right-angled, so they have two pairs of congruent angles.

If we choose $B$ so that $OA$=$OB,$ then triangles $△OBD$ and $△OAC$ also have a congruent side. According to the Angle-Angle-Side (AAS) Congruence Theorem, the triangles are congruent and hence all corresponding sides are congruent. Let's see how we can use this to find the coordinates of $B$ and $D.$

• Point $D$ is on the $x\text{-}$axis, so the $y\text{-}$coordinate is $0.$ Since $AC=b$ and $\overline{DO}$ is congruent to $\overline{AC},$ the $x\text{-}$coordinate of $D$ is $\text{-}b.$
• Segment $\overline{BD}$ is vertical, so the $x\text{-}$coordinate of $B$ is also $\text{-}b.$ Since $OC=a$ and $\overline{DB}$ is congruent to $\overline{OC},$ the $y\text{-}$coordinate of $B$ is $a.$

Let's use the Slope Formula to find the slope of the lines.

Line	Points	Slope $\left(\frac{y_2-y_1}{x_2-x_1}\right)$
		Substitution	Simplification
${\ell }_1$	$O(0,0)$ and $A(a,b)$	$\frac{b-0}{a-0}$	$m_1=\frac{b}{a}$
${\ell }_2$	$O(0,0)$ and $B(\text{-}b,a)$	$\frac{a-0}{\text{-}b-0}$	$m_2=\frac{a}{\text{-}b}=\text{-}\frac{a}{b}$

Let's find the product of the slopes. The product of the slopes of perpendicular lines is indeed $\text{-}1.$