Lesson 1: Lines Review

Is the point $(1,-2)$ a solution to the equation $x-2y=3$?

In $(1,-2)$ we have $x=1$ and $y=-2$.

\begin{align*} x-2y &\stackrel{?}{=}3\\ 1-2(-2)&\stackrel{?}{=}3\\ 1+4 &\stackrel{?}{=}3\\ 5&\stackrel{\times}{=}3\qquad\text{False} \end{align*}

The point $(1,-2)$ is not a solution to the equation $x-2y=3$.

Isolate $y$.

$$2x-3y=6$$

\begin{align*}2x-3y&=6\\-3y&=-2x+6\\y&=\dfrac{2}{3}x-2\end{align*}

Write an equation of the graphed line using the point-slope form.

The line passes through $(x_1,y_1)=(-2,0)$ and $(x_2,y_2)=(3,-2)$. The slope is

\begin{align*}m &= \dfrac{y_2-y_1}{x_2-x_1} \\ &= \dfrac{-2-0}{3-(-2)}\\& = \dfrac{-2}{5}\\ &=-\dfrac{2}{5}.\end{align*}

The point-slope form is

\begin{align*} y-y_1&=m(x-x_1)\\ y-0&=-\dfrac{2}{5}(x-(-2))\\ y&=-\dfrac{2}{5}(x+2)\end{align*}

or

\begin{align*}y-y_2&=m(x-x_2)\\y-(-2)&=-\dfrac{2}{5}(x-3)\\ y+2&=-\dfrac{2}{5}(x-3).\end{align*}

Draw a line with equation $$y-3=-\dfrac{1}{2}(x+1).$$

By setting $x=0$, we obtain that $y=5/2$, so the $y$-intercept is $(0,5/2)$.

By setting $y=0$, we obtain that $x=5$, so the $x$-intercept is $(5,0)$.

Write the equation for the graph below in slope-intercept form.

The line passes through $(x_1,y_1)=(0,-1)$ and $(x_2,y_2)=(3,2)$. The slope is

\begin{align*}m &= \dfrac{y_2-y_1}{x_2-x_1}\\ &= \dfrac{2-(-1)}{3-0}\\ &= \dfrac{3}{3}\\&=1.\end{align*}

The $y$-intercept is $(0,-1)$, so $b=-1$. The slope-intercept form is

\begin{align*}y&=mx+b\\y&=1\cdot x -1\\y&=x-1\end{align*}

Graph $y=2x-1$.

When $y=2x-1$, the slope is $2$, the $y$-intercept is $(0,-1)$, and the $x$-intercept is $(1/2,0)$.

Graph $y=-3$.

This is a horizontal line passing through $(0,-3)$.

Write an equation in standard form for a line passing through $(1,3)$ and $(-5,2)$.

The line passes through $(x_1,y_1)=(1,3)$ and $(x_2,y_2)=(-5,2)$. The slope is

\begin{align*}m &= \dfrac{y_2-y_1}{x_2-x_1}\\&= \dfrac{2-3}{-5-1}\\& = \dfrac{-1}{-6}\\&=\dfrac{1}{6}.\end{align*}

The equation of the line is

\begin{align*}y-y_1&=m(x-x_1)\\ y-3&=\dfrac{1}{6}(x-1)\\ y-3&=\dfrac{1}{6}x-\dfrac{1}{6}\\ -\dfrac{1}{6}x+y&= 3-\dfrac{1}{6}\\ -\dfrac{1}{6}x+y&= \dfrac{17}{6}\\x-6y&=- 17\end{align*}

Graph $y+2x=4$ by plotting points.

When $x=1$, $y=2$, so $(1,2)$ is a point on the line.

When $x=-3$, $y=10$, so $(-3,10)$ is a point on the line.

• Explain the notion of the slope and give reasoning behind the slope formula.
• How do you write a given linear equation in slope-intercept form?
• If two lines are perpendicular/parallel, what is the relationship between their slopes?
• Explain reasoning behind the midpoint formula.
• How do you graph the solutions to a linear equation in two variables if it is given in point-slope form?
• How does the information you can see from the graph inform your production of an equation whose solution is that graph?
• If a point $(2,3)$ is not on the graph of an equation, what can you say about the equation?

$\bigstar$ Write an equation for a line perpendicular to $5x-2y=6$ passing through $(-1,2)$.

\begin{align*}5x-2y&=6\\-2y&=-5x+6\\y&=\dfrac{5}{2}x-3\end{align*}

The slope of the line $5x-2y=6$ is $\dfrac{5}{2}$. The slope of the perpendicular line is $-\dfrac{2}{5}$. The equation of the line with slope $-\dfrac{2}{5}$ passing through $(-1,2)$ is

\begin{align*}y-2&=-\dfrac{2}{5}(x-(-1))\\y-2&=-\dfrac{2}{5}(x+1)\end{align*}