# Write the equation of the line, in standard form, that passes through the points (2, -3) and (-1,...

## Question:

Write the equation of the line, in standard form, that passes through the points (2, -3) and (-1, -4).

## Point-Slope Form and Slope-Intercept Form:

(i) The point-slope form of a line of slope {eq}m {/eq} that passes through a point {eq}(x_1,y_1) {/eq} is given by:

$$y-y_1=m(x-x_1) $$

(ii) The slope-intercept form of a line of slope {eq}m {/eq} and y-intercept {eq}b {/eq} is given by:

$$y=mx+b $$

The equation of point-slope form can be converted to the equation of slope-intercept form using algebraic operations.

## Answer and Explanation:

The points on the line are:

$$(x_1,y_1)=(2,-3)\\ (x_2, y_2)= (-1,-4) $$

The slope of the line is found using:

$$\begin{align} m&= \dfrac{y_2-y_1}{x_2-x_1} \\ &= \dfrac{-4-(-3)}{-1-2} \\ &= \dfrac{-1}{-3} \\ &= \dfrac{1}{3} \end{align} $$

The equation of the line is found using the point-slope form:

$$y-y_1=m(x-x_1)\\ y-(-3)=\dfrac{1}{3} (x-2) \\ y+3=\dfrac{1}{3} (x-2) \\ y+3=\dfrac{1}{3} - \dfrac{2}{3} \\ \text{Subtracting 3 from both sides}, \\ y=\dfrac{1}{3}x -\dfrac{2}{3}-3 \\ \boxed{\mathbf{y= \dfrac{1}{3} x - \dfrac{11}{3}}} $$.

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