Question

1. examine the graph of line $overleftrightarrow{cm}$ at right. homework help

a. write the equation of $overleftrightarrow{cm}$.
b. what are the $x$- and $y$-intercepts of the line?
c. calculate the area and perimeter of $\triangle cpm$.
d. write an equation of the line through point $m$ that is perpendicular to $overleftrightarrow{cm}$.

Explanation:

Step1: Find the slope of line $\overleftrightarrow{CM}$

The two - point form of the slope $m=\frac{y_2 - y_1}{x_2 - x_1}$. From the graph, $C(-2,1)$ and $M(2,3)$. So $m=\frac{3 - 1}{2-(-2)}=\frac{2}{4}=\frac{1}{2}$. Using the point - slope form $y - y_1=m(x - x_1)$ with the point $C(-2,1)$, we have $y - 1=\frac{1}{2}(x + 2)$. Simplifying gives $y=\frac{1}{2}x+2$.

Step2: Find the $x$ and $y$ intercepts

For the $y$ - intercept, set $x = 0$ in $y=\frac{1}{2}x + 2$, we get $y=2$. For the $x$ - intercept, set $y = 0$ in $y=\frac{1}{2}x+2$, then $0=\frac{1}{2}x + 2$, which gives $x=-4$.

Step3: Calculate the side lengths of $\triangle CPM$

$CP$: The $y$ - coordinate of $C$ is $1$ and of $P$ is $1$, and $x$ - coordinate of $C$ is $-2$ and of $P$ is $2$. So $CP=\vert2-(-2)\vert = 4$. $PM$: The $x$ - coordinate of $P$ is $2$ and of $M$ is $2$, and $y$ - coordinate of $P$ is $1$ and of $M$ is $3$. So $PM=\vert3 - 1\vert=2$. Using the Pythagorean theorem, $CM=\sqrt{(2 + 2)^2+(3 - 1)^2}=\sqrt{16 + 4}=\sqrt{20}=2\sqrt{5}$. The area of $\triangle CPM$ is $A=\frac{1}{2}\times CP\times PM=\frac{1}{2}\times4\times2 = 4$. The perimeter is $P=4 + 2+2\sqrt{5}=6 + 2\sqrt{5}$.

Step4: Find the equation of the perpendicular line

The slope of a line perpendicular to a line with slope $m$ is $m'=-2$ (since $m\times m'=-1$ for perpendicular lines). Using the point - slope form with the point $M(2,3)$, we have $y - 3=-2(x - 2)$. Simplifying gives $y=-2x + 7$.

Answer:

a. $y=\frac{1}{2}x + 2$
b. $x$ - intercept: $x=-4$, $y$ - intercept: $y = 2$
c. Area: $4$, Perimeter: $6 + 2\sqrt{5}$
d. $y=-2x + 7$