Section 1.3 – Linear Equations in Two Variables
- 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
Exercises
Directions: You should try to solve each problem first, and then click "Reveal Answer" to check your answer. You can click "Watch Video" if you need help with a problem.
1. Find the 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 \(y=\dfrac{2}{5}x+2\).
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 \(x=5\).
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
\[y-2=-\dfrac{2}{3}(x-3)\]
4. Write the equation of the line parallel \(5x-4y=8\) passing through the point \((3,-2)\).
Slope-Intercept Form: \(\displaystyle y=\frac{5}{4}x-\frac{23}{4}\)
Standard Form: \(\displaystyle 5x-4y=23\)