CPI Precalculus: Extra Practice on Linear Equations Worksheet 2
- Class: CPI Precalculus
- Author: Peter Atlas
- Text: Advanced Mathematics, Brown
- Find an equation for the line described:
- passes through (-3, -1) and has a slope 4.
Solution
\( y + 1 = 4(x + 3) \)
- passes through \( \displaystyle \left( \frac{5}{2}, 0 \right) \) and has a slope \( \displaystyle \frac{1}{2}.\)
Solution
\( \displaystyle y = \frac{1}{2} \left( x - \frac{5}{2} \right) \)
- has x-intercept 6 and y-intercept 5
Solution
\( \displaystyle \frac{x}{6} + \frac{y}{5} = 1 \)
- has x-intercept -2 and slope \( \displaystyle \frac{3}{4}.\)
Solution
\( \displaystyle y + 2 = \frac{3}{4}x \)
- passes through (1, 2) and (2, 6).
Solution
\( y - 2 = 4(x - 1) \)
- passes through (-7, -2) and (0, 0).
Solution
\( \displaystyle y = \frac{2x}{7} \)
- passes through (-3, 4) and is parallel to the x-axis.
Solution
\( y = 4 \)
- passes through (-3, 4) and is parallel to the y-axis.
Solution
\( x = -3 \)
- Determine whether each pair of lines is parallel, perpendicular, or neither
- \(3x - 4y = 12\) and \(4x - 3y = 12\)
Solution
Neither
- \(y = 5x - 16\) and \( y = 5x + 2\)
Solution
Parallel
- \( 5x - 6y = 25\) and \( 6x + 5y = 0\)
Solution
Perpendicular
- \(x = 8y + 3\) and \( \displaystyle 4y - \frac{x}{2} = 32\)
Solution
Parallel
- Find the equation of the line described. Write your \( ANSWER \) in standard form:
- is parallel to \(2x - 5y = 10\) and passes through (-1, 2).
Solution
\( 2x - 5y = -12 \)
- is perpendicular to \(4y - 3x = 1\) and passes through (4, 0).
Solution
\( 4x + 3y = 16 \)
- is perpendicular to \(x - y + 2 = 0\) and passes through (3, 1).
Solution
\( x + y = 4 \)
- is parallel to \(3x - 5y = 25\) and has the same y-intercept as the line \(6x - y + 11 = 0\).
Solution
\( 3x - 5y = -55 \)
- Find the slope of the line passing through the two points (3, 9) and \( (3 + h, (3 + h)^2)\).
Solution
\( 6 + h \)
- Find the slope of the line passing through the two points \( (x, x^2) \) and \( (x + h, (x + h)^2) \).
Solution
\( 2x + h \)
- Find the linear function satisfying the given conditions :
- \(f(-1) = 0\) and \(f(5) = 4\).
Solution
\( \displaystyle y = \frac{2}{3}(x + 1) \)
- f(3) = 2 and f(-3) = -4
Solution
\( y - 2 = x - 3 \)
- \(g(0) = 0\) and \(g(1) = \sqrt{2}.\)
Solution
\( y = \sqrt{2}x \)
- The graph passes through the points (2, 4) and (3, 9).
Solution
\( y - 4 = 5(x - 2) \)
- \( \displaystyle f \left( \frac{1}{2} \right) = -3\), and the graph of \(f\) is a line parallel to the line \(x - y = 1\).
Solution
\( \displaystyle y + 3 = x - \frac{1}{2} \)
- \(g(2) = 1\) and the graph of \(g\) is perpendicular to the line \(6x - 3y = 2\)
Solution
\( \displaystyle y - 1 = -\frac{1}{2}(x - 2) \)