Chapter 5.3, Problem 23PSB

To determine

To Find: Which of the given points does lies on the given line.

Answer to Problem 23PSB

  $\begin{array}{l}\text{Line }y=5x-7\text{ is passing through the points }\\ A=\left(3,8\right),B=\left(-2,-17\right),C=\left(0,-7\right)\text{ and }E=\left(1,-2\right)\end{array}$

Explanation of Solution

Given:

  $\begin{array}{l}\text{Equation of line }y=5x-7\\ A=\left(3,8\right),B=\left(-2,-17\right),C=\left(0,-7\right),D=\left(4,11\right)\text{ and }E=\left(1,-2\right)\end{array}$

Concept Used:

Any point would lies on the given line if it satisfy the given equation of line.

Calculation:

  $\begin{array}{l}\text{Equation of line }y=5x-7\left(1\right)\\ A=\left(3,8\right),B=\left(-2,-17\right),C=\left(0,-7\right),D=\left(4,11\right)\text{ and }E=\left(1,-2\right)\\ \text{For }A=\left(3,8\right),y=5\times 3-7=8\\ \Rightarrow \text{Line}\left(1\right)\text{is passing through }A=\left(3,8\right)\\ \text{For }B=\left(-2,-17\right),y=5\times \left(-2\right)-7=-17\\ \Rightarrow \text{Line}\left(1\right)\text{is passing through }B=\left(-2,-17\right)\\ \text{For }C=\left(0,-7\right),y=5\times 0-7=-7\\ \Rightarrow \text{Line}\left(1\right)\text{is passing through }C=\left(0,-7\right)\\ \text{For }D=\left(4,11\right),y=5\times 4-7=13\ne 11\\ \Rightarrow \text{Line}\left(1\right)\text{is not passing through }D=\left(4,11\right)\\ \text{For }E=\left(1,-2\right),y=5\times 1-7=-2\\ \Rightarrow \text{Line}\left(1\right)\text{is passing through }E=\left(1,-2\right)\end{array}$

  $\begin{array}{l}\text{Slope of line joining }A=\left(1,5\right)\text{and }B=\left(4,9\right)\text{is}{m}_1=\frac{9-5}{4-1}=\frac{4}{3}\left(1\right)\\ \text{Slope of line joining }C=\left(6,0\right)\text{ and }D=\left(9,4\right)\text{is}{m}_2=\frac{4-0}{9-6}=\frac{4}{3}\left(2\right)\\ \because m_1=m_2\\ \text{Since, we know that when two lines are parallel, their slope are also equal}\text{.}\\ \Rightarrow \overline{AB}\parallel \overline{CD}\\ \Rightarrow \overleftrightarrow{m}\parallel \overleftrightarrow{n}\end{array}$