## Slope of Perpendicular Lines Theorem

If you have already learned about systems of linear equations, then you have probably discussed that the product of the slope of perpendicular lines is $-1$. The proof of this theorem comes from the fact that any point $(x,y)$ rotated 90 degrees at about the origin becomes $(-y,x)$. One example of this is shown below. The point $(3, 4)$, when rotated $90$ degrees counterclockwise becomes $(-4,3)$.

With this fact, we prove this theorem.

Slope of Perpendicular Lines Theorem

If two lines with slopes $m_1$ and $m_2$ are perpendicular, then $m_1 m_2 = -1$Continue reading…

## The Sum and Product of Roots Theorem

From the quadratic formula, we know that the numbers $r_1$ and $r_2$ are the roots of the quadratic equation $ax^2 + bx +c =0$ where $a \neq 0$ if and only if

$r_1 + r_2 = -\frac{b}{a}$

and

$r_1r_2 = \frac{c}{a}$Continue reading…

## Sample Proof on Triangle Congruence Part 3

Recall that for real numbers a, b, and c are real numbers, if $a = b$ and $b = c$, then $a = c$. This is called Transitive Property of Equality. This is also the same with congruence. If $A$, $B$, and $C$ are polygons, and if $A$ is congruent to $B$, and $B$ is congruent $C$, then $A$ is congruent to $C$. This is called the Transitive Property of Congruence. We will use this to prove the following problem.

Given: $\overline{BE} \cong \overline{DC}$ and $\overline{BD} \cong \overline{CA}$.

Prove: $\triangle DBE \cong \triangle CAB$.

Proof

It is given that $\overline{BE} \cong \overline{DC}$.

Now, by reflexive property, that is a segment is congruent to itself, $\overline{BD} \cong \overline{BD}$Continue reading…