Question

in this coordinate plane, the slope of line m is equal to the opposite reciprocal of the slope of line n. complete the proof that line m is perpendicular to line n.

Explanation:

Step1: Recall slope - formula

The slope formula for a line passing through two points $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2 - y_1}{x_2 - x_1}$.

Step2: Find slope of line $m$

The line $m$ passes through points $A(0,0)$ and $C(b,c)$. Using the slope - formula, the slope of line $m$, denoted as $m_1=\frac{c - 0}{b - 0}=\frac{c}{b}$ ($b
eq0$).

Step3: Find slope of line $n$

The line $n$ passes through points $A(0,0)$ and $D(b,-d)$. Using the slope - formula, the slope of line $n$, denoted as $m_2=\frac{-d - 0}{b - 0}=-\frac{d}{b}$ ($b
eq0$).

Step4: Use given relationship

We know that the slope of line $m$ is the opposite reciprocal of the slope of line $n$, i.e., $m_1=-\frac{1}{m_2}$. Substituting $m_1=\frac{c}{b}$ and $m_2 =-\frac{d}{b}$, we get $\frac{c}{b}=-\frac{1}{-\frac{d}{b}}=\frac{b}{d}$ (assuming $d
eq0$). Cross - multiplying gives $cd = b^2$.

Step5: Use vector or geometric approach

Let $\overrightarrow{AC}=(b,c)$ and $\overrightarrow{AD}=(b,-d)$. The dot - product of two vectors $\overrightarrow{u}=(x_1,y_1)$ and $\overrightarrow{v}=(x_2,y_2)$ is $\overrightarrow{u}\cdot\overrightarrow{v}=x_1x_2 + y_1y_2$. Here, $\overrightarrow{AC}\cdot\overrightarrow{AD}=b\times b + c\times(-d)=b^2 - cd$. Since $cd = b^2$, then $\overrightarrow{AC}\cdot\overrightarrow{AD}=0$.
If the dot - product of two non - zero vectors is zero, the vectors are perpendicular. Since $\overrightarrow{AC}$ lies on line $m$ and $\overrightarrow{AD}$ lies on line $n$, line $m$ is perpendicular to line $n$.

Answer:

Line $m$ is perpendicular to line $n$ because the dot - product of vectors lying on them is zero, which is a consequence of the given relationship between their slopes.