Chapter 1, Problem 125RE

To determine

To calculate:The equation for the line which has the value of $x$ -intercept as $4$ and that of $y$ −intercept as $12$ .

Answer to Problem 125RE

The equation for the line which has the value of $x$ -intercept as $4$ and that of $y$ −intercept as $12$ is $y+3x-12=0$

Explanation of Solution

Given information:

For a given line the value of $x$ -intercept as $4$ and that of $y$ −intercept as $12$ .

Formula used:

Slope m of the line passing through two points in general say $P_1=\left(x_1,y_1\right)$ and $P_2=\left(x_2,y_2\right)$ is:

  $m=\frac{y_2-y_1}{x_2-x_1}$

Point-slope equation for a given line passing along point $P_1=\left(x_1,y_1\right)$ is:

  $y-y_1=m\left(x-x_1\right)$

Slope-intercept equation for a given line which has slope as $m$ and $y$ −intercept as $b$ is:

  $y=mx+b$

Two-intercept equation for a given line which has $x$ -intercept as $a$ and $y$ −intercept as $b$ is:

  $\frac{x}{a}+\frac{y}{b}=1$

Calculation:

Consider thegiven values of $x$ -intercept as $4$ and that of $y$ −intercept as $12$ .

Then in order to proceed following should be known:

  $\begin{array}{l}a=4\\ b=12\end{array}$

Recall the two-intercept equation for a given line which has $x$ -intercept as $a$ and $y$ −intercept as $b$ is:

  $\frac{x}{a}+\frac{y}{b}=1$

Putting the values to get:

  $\begin{array}{l}\frac{x}{4}+\frac{y}{12}=1\\ \frac{3x+y}{12}=1\\ 3x+y=12\\ y+3x-12=0\end{array}$

Hence, the equation for the line which has the value of $x$ -intercept as $4$ and that of $y$ −intercept as $12$ is $y+3x-12=0$