Chapter 1.1, Problem 62E

To determine

the equation of the line that passes through the given points using slope intercept form if possible and plot graph.

Answer to Problem 62E

  $3x+25y-318=0$

Explanation of Solution

Given:

Given point is $\left(\frac{3}{4},\frac{3}{2}\right)\text{and}\left(-\frac{4}{3},\frac{7}{4}\right)$ .

Calculation:

Standard equation for slope intercept form is.

  $y=mx+b$

Here, is $m$ the slope of line and $b$ is the y-intercept.

Calculating $m$ for the given coordinates.

  $m=\frac{y_2-y_1}{x_2-x_1}$

Substituting $x_1=\frac{3}{4},y_1=\frac{3}{2},x_2=-\frac{4}{3}\text{and }{y}_2=\frac{7}{4}$ in the above equation.

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{\frac{7}{4}-\frac{3}{2}}{-\frac{4}{3}-\frac{3}{4}}\\ m=\frac{\frac{1}{4}}{-\frac{25}{12}}\\ m=-\frac{3}{25}\end{array}$

Now substituting $m=-\frac{3}{25}$ and $x=\frac{3}{4},y=\frac{3}{2}$ in equation $y=mx+b$ .

  $\begin{array}{l}y=mx+b\\ \frac{3}{2}=\left(-\frac{3}{25}\right)\times \frac{3}{4}+b\\ \frac{3}{2}+\frac{9}{100}=b\\ \frac{318}{200}=b\end{array}$

Put $\frac{318}{200}=b$ and $m=-\frac{3}{25}$ in equation $y=mx+b$ .

  $\begin{array}{l}y=mx+b\\ y=\left(-\frac{3}{25}\right)x+\frac{318}{25}\\ y=-\frac{3}{25}x+\frac{318}{25}\\ 25y=-3x+318\\ 3x+25y-318=0\end{array}$

Therefore, required equation is $3x+25y-318=0$ .

Graph is drawn as for coordinates $\left(0,12.72\right)\text{and }\left(106,0\right)$ .

  

Therefore, the answer is $3x+25y-318=0$ .