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Section 1.3 – Linear Equations in Two Variables

Section Details.
  • The general equation of a line and finding the slope of a line
  • Equation for a line including the slope-intercept form and point-slope form
  • Finding the equation of a line and graphing a line
  • Equations and slopes for horizontal and vertical lines
  • Parallel and Perpendicular lines


Practice Problems


Directions. The following are review problems for this section. It is recommended that you work the problems, and then click "Solution" to check your answer. If you do not understand how to solve a problem, you can click "Video" to learn how to solve it. 
  1. Find an equation of the line through the points \((5,4)\) and \((-10,-2)\). 

    Answer: \(y=\dfrac{2}{5}x+2\)

    Solution: Slope \(m=\dfrac{4-(-2)}{5-(-10)}=\dfrac{6}{15}=\dfrac{2}{5}\)

    \(y-4=\dfrac{2}{5}(x-5)\)  OR  \(y-(-2)=\dfrac{2}{5}(x-(-10))\)

    Simplifies to \(y=\dfrac{2}{5}x+2\).


    To see the full video page and find related videos, click the following link.
    WIR1 20B M150 V1


  2. Find an equation of the line through the points \((5,4)\) and \((5,-2)\). 

    Answer: \(x=5\)

    Solution:
    Slope \(m=\dfrac{4-(-2)}{5-5}=\dfrac{6}{0}\) which is undefined

    Undefined slope means the line is vertical and all x values are the same, so our equation is \(x=5\).


    To see the full video page and find related videos, click the following link.
    WIR1 20B M150 V2


  3. Write an equation of a line a) parallel to and b) perpendicular to the line \(5+6x-4y=0\) and passing through the point \((3,2)\). 

    Answer: a) \(y-2=\dfrac{3}{2}(x-3)\), b) \(y-2=-\dfrac{2}{3}(x-3)\)

    Solution: First, we need to know the slope of our given line.

    \[5+6x-4y=0\]

    \[5+6x=4y\]

    \[\dfrac{5}{4}+\dfrac{3}{2}x=y\]

    a) The slope of the given line is \(\dfrac{3}{2}\), so the parallel line for a) will also have a slope of \(m=\dfrac{3}{2}\).

    \[y-2=\dfrac{3}{2}(x-3)\]

    b) The perpendicular slope will be \(m_{\perp}=-\dfrac{2}{3}\), so the equation of the perpendicular line will be

    \[y-2=-\dfrac{2}{3}(x-3)\]


    To see the full video page and find related videos, click the following link.
    WIR1 20B M150 V3


  4. Write the equation of the line parallel \(5x-4y=8\) passing through the point \((3,-2)\).

    Point-Slope Form: \(\displaystyle y-(-2)=\frac{5}{4}(x-3)\)
    Slope-Intercept Form: \(\displaystyle y=\frac{5}{4}x-\frac{23}{4}\)
    Standard Form: \(\displaystyle 5x-4y=23\)


    To see the full video page and find related videos, click the following link.
    WIR8 20B M150 V7