Chapter 4.2, Problem 36PPS

To determine

To find: The equation of the straight line through the pair of points $\left(\frac{5}{4},1\right)$ and $\left(-\frac{1}{4},\frac{3}{4}\right)$ .

Answer to Problem 36PPS

The equation of straight line is $4x-24y+19=0$ .

Explanation of Solution

Given information: The given two pair of points through which the line passes are $\left(\frac{5}{4},1\right)$ and $\left(-\frac{1}{4},\frac{3}{4}\right)$ .

Formula used: Using two point form of the line, the equation of the line is written as

  $y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$

Calculation: Using the two point form of the line, the equation of the line is written as,

  $\begin{array}{c}y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)\mathrm{......}\left(\begin{array}{l}\left(x_1,y_1\right)\to \left(\frac{5}{4},1\right)\\ \left(x_2,y_2\right)\to \left(-\frac{1}{4},\frac{3}{4}\right)\end{array}\right)\\ y-1=\frac{\frac{3}{4}-1}{-\frac{1}{4}-\frac{5}{4}}\left(x-\frac{5}{4}\right)\\ y-1=\frac{\left(-\frac{1}{4}\right)}{-\left(\frac{6}{4}\right)}\left(x-\frac{5}{4}\right)\mathrm{......}\left(\text{Simplifying}\right)\\ y-1=\frac{1}{6}\left(x-\frac{5}{4}\right)\mathrm{......}\left(1\right)\end{array}$

Further solving (1), the equation of the straight line is written as

  $\begin{array}{c}6\left(y-1\right)=\left(x-\frac{5}{4}\right)\mathrm{......}\left(\text{Cross multiplying}\right)\\ 6y-6=x-\frac{5}{4}\\ 4\left(6y-6\right)=4\left(x-\frac{5}{4}\right)\mathrm{......}\left(\text{Multiplying both sides by 4}\right)\\ 24y-24=4x-5\\ \text{or, }4x-24y+19=0\end{array}$

Hence, the required equation of the straight line through the points $\left(\frac{5}{4},1\right)$ and $\left(-\frac{1}{4},\frac{3}{4}\right)$ is $4x-24y+19=0$ .