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11) \( (-7,9),(8,1) \) Find the slope of each line. Also, find the \( y \)-intercept. \( \begin{array}{ll}\text { 13) } y=-\frac{5}{3} x & \text { 12) }(-8,12),(-8,-14) \\ \text { Sketch the graph of each line on separate graph paper using the slope-intercept method! } \\ \text { 15) } y=-\frac{8}{5} x-5 & \text { 14) } y=-\frac{1}{4} x+4\end{array} \) 16) \( y=2 x-3 \)

Ask by Moreno Black. in the United States
Mar 09,2025

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**Simplified Answers:** 1. **Problem 11:** - **Slope (\( m \))**: \(-\frac{8}{15}\) - **\( y \)-intercept**: \(\left(0, \frac{79}{15}\right)\) 2. **Problem 12:** - **Slope**: Undefined (vertical line) - **\( y \)-intercept**: None (equation is \( x = -8 \)) 3. **Problem 13:** - **Slope (\( m \))**: \(-\frac{5}{3}\) - **\( y \)-intercept**: \((0, 0)\) 4. **Problem 14:** - **Slope (\( m \))**: \(-\frac{1}{4}\) - **\( y \)-intercept**: \((0, 4)\) 5. **Problem 15:** - **Slope (\( m \))**: \(-\frac{8}{5}\) - **\( y \)-intercept**: \((0, -5)\) 6. **Problem 16:** - **Slope (\( m \))**: \(2\) - **\( y \)-intercept**: \((0, -3)\)

Solution

**Problem 11: Line Through \((-7,9)\) and \((8,1)\)** 1. To find the slope \( m \), use the formula \[ m = \frac{y_2-y_1}{x_2-x_1}. \] 2. Substitute the given points: \[ m = \frac{1-9}{8-(-7)} = \frac{-8}{15} = -\frac{8}{15}. \] 3. To find the \( y \)-intercept, use the point-slope form: \[ y - y_1 = m(x - x_1). \] Using point \((-7,9)\): \[ y - 9 = -\frac{8}{15}(x - (-7)) = -\frac{8}{15}(x+7). \] 4. Substitute \( x=0 \) to find the \( y \)-intercept: \[ y - 9 = -\frac{8}{15}(0+7) = -\frac{56}{15}. \] \[ y = 9 - \frac{56}{15}. \] 5. Express \( 9 \) as a fraction with denominator 15: \[ 9=\frac{135}{15}\quad\Longrightarrow\quad y=\frac{135}{15}-\frac{56}{15}=\frac{79}{15}. \] 6. The \( y \)-intercept is \(\left(0, \frac{79}{15}\right)\). --- **Problem 12: Line Through \((-8,12)\) and \((-8,-14)\)** 1. Use the slope formula: \[ m = \frac{-14-12}{-8-(-8)} = \frac{-26}{0}. \] 2. Since division by zero is undefined, the slope is undefined. 3. The line is vertical with equation \( x=-8 \). 4. A vertical line does not have a \( y \)-intercept (unless \( x=-8 \) equals 0, which it does not). --- **Problem 13: \( y=-\frac{5}{3}x \)** 1. The equation is in slope-intercept form \( y=mx+b \). 2. The slope is: \[ m=-\frac{5}{3}. \] 3. The \( y \)-intercept is at: \[ b=0,\quad \text{or} \quad(0,0). \] --- **Problem 14: \( y=-\frac{1}{4}x+4 \)** 1. The equation is in slope-intercept form. 2. The slope is: \[ m=-\frac{1}{4}. \] 3. The \( y \)-intercept is: \[ b=4,\quad \text{or} \quad(0,4). \] --- **Problem 15: \( y=-\frac{8}{5}x-5 \)** 1. The equation is already in slope-intercept form. 2. The slope is: \[ m=-\frac{8}{5}. \] 3. The \( y \)-intercept is: \[ b=-5,\quad \text{or} \quad(0,-5). \] --- **Problem 16: \( y=2x-3 \)** 1. The equation is in slope-intercept form. 2. The slope is: \[ m=2. \] 3. The \( y \)-intercept is: \[ b=-3,\quad \text{or} \quad(0,-3). \]

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To find the slope of the line through the points \((-7,9)\) and \((8,1)\), use the formula for slope \( m = \frac{y_2 - y_1}{x_2 - x_1} \). Plugging in the coordinates gives \( m = \frac{1 - 9}{8 - (-7)} = \frac{-8}{15} \). The \( y \)-intercept can be found by plugging one of the points into the line equation \( y = mx + b \). Using point \((8, 1)\): \( 1 = -\frac{8}{15}(8) + b \) results in the \( y \)-intercept \( b = \frac{41}{15} \). For the second pair \((-8,12)\) and \((-8,-14)\), since the x-coordinates are the same, the line is vertical, therefore, it doesn’t have a slope and \( x = -8 \) is the equation for this vertical line. When sketching these lines using the slope-intercept method, remember that the slope tells you how to rise over run, which creates the incline of your line. Use your \( y \)-intercept to start plotting your point on the graph, and then from that point move according to the slope to find another point. Repeat for the other equations, ensuring to draw them on separate pieces of graph paper to keep things neat!

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