Chapter 1.1, Problem 61E

To determine

To find: the equation of the line that passes through the point $\left(2,\frac{1}{2}\right),\left(\frac{1}{2},\frac{5}{4}\right)$ using slope intercept form if possible and draw graph.

Answer to Problem 61E

  $x-2y-1=0$

Explanation of Solution

Given:

Given point are $\left(2,\frac{1}{2}\right),\left(\frac{1}{2},\frac{5}{4}\right)$ .

Calculation:

Standard equation for slope-intercept form is.

  $y=mx+b$

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

Slope is calculated as.

  $m=\frac{y_2-y_1}{x_2-x_1}$

Substituting $x_1=2,y_1=\frac{1}{2},x_2=\frac{1}{2}\text{and }{y}_2=\frac{5}{4}$ .

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{\frac{5}{4}-\frac{1}{2}}{\frac{1}{2}-2}\\ m=\frac{\frac{-3}{4}}{-\frac{3}{2}}\\ m=\frac{1}{2}\end{array}$

Now, substituting $x=2,y=\frac{1}{2},m=\frac{1}{2}$ in equation $y=mx+b$ .

  $\begin{array}{l}y=mx+b\\ \frac{1}{2}=\frac{1}{2}\times 2+b\\ \frac{1}{2}-1=b\\ b=-\frac{1}{2}\end{array}$

Now, substituting $m=\frac{1}{2}\text{ and }b=-\frac{1}{2}$ in equation $y=mx+b$ .

  $\begin{array}{l}y=mx+b\\ y=\frac{1}{2}x+\left(-\frac{1}{2}\right)\\ y=\frac{1}{2}x-\frac{1}{2}\\ 2y=x-1\\ x-2y-1=0\end{array}$

Graph is drawn for coordinates as.

  $\begin{array}{cc}x& y\\ 0& -0.5\\ 1& 0\end{array}$

  

As it is clearly visible from the graph that it is straight line passing through $y=-0.5$ and $x=1$ .

Therefore, the answer is $x-2y-1=0$ .