Q9. - Algebra 1

Question

Write an equation of the line with the given conditions.

$\left(-1,6\right),\left(2,4\right)$.

Step-by-Step Solution

Answer

The equation of the straight line for the given condition is $\mathit{y}\mathbf{=}\frac{\mathbf{-}\mathbf{2}}{\mathbf{3}}\mathit{x}\mathbf{+}\frac{\mathbf{16}}{\mathbf{3}}$.

Step 1. State the concept of a straight line.

The standard form of slope and intercept form of the straight line is $y=mx+c$.

Here, $m$ is the slope of the straight line and $c$ is the $y$-intercept of the equation.

Step 2. State the equation of a line slope-point form.

The standard form of the two-point form of the straight line is $y-y_1=\left(\frac{y_2-y_1}{x_2-x_1}\right)\left(x-x_1\right)$.

Here, coordinates of two points from which the straight line passes.

$\begin{array}{l}\left(x_1,y_1\right)=\left(-1,6\right)\\ \left(x_2,y_2\right)=\left(2,4\right)\end{array}$

Step 3. Substitute the values to get the required equation.

Substitute the above values in the equation $y-y_1=\left(\frac{y_2-y_1}{x_2-x_1}\right)\left(x-x_1\right)$.

$$\begin{gathered} y-y_1=\left(\frac{y_2-y_1}{x_2-x_1}\right)\left(x-x_1\right) \\ y-6=\left(\frac{4-6}{2-\left(-1\right)}\right)\left(x-\left(-1\right)\right) \\ y-6=\frac{-2}{3}\left(x+1\right) \\ y=\frac{-2}{3}x-\frac{2}{3}+6 \\ y=\frac{-2}{3}x+\frac{16}{3} \end{gathered}$$

The equation of the straight line is $y=\frac{-2}{3}x+\frac{16}{3}$.