How do you write the equation of a line in slope intercept, point slope and standard form given (2,-3) and (4,-1)?

Apr 17, 2017

The equation of the line in slope-intercept form is $y = x - 5$
in point $\left(2 , - 3\right)$ slope form is $y + 3 = x - 2$
and in standard form is $- x + y = - 5$

Explanation:

The slope of the line passing through $\left(2 , - 3\right) \mathmr{and} \left(4 , - 1\right)$ is $m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{- 1 + 3}{4 - 2} = \frac{2}{2} = 1$

Let the equation of the line in slope-intercept form be $y = m x + c \mathmr{and} y = x + c$ The point (2,-3) will satisfy the equation . So, $- 3 = 2 + c \mathmr{and} c = - 5$

Hence the equation of the line in slope-intercept form is $y = x - 5$

The equation of the line in point $\left(2 , - 3\right)$ slope form is $y - {y}_{1} = m \left(x - {x}_{1}\right) \mathmr{and} y - \left(- 3\right) = 1 \cdot \left(x - 2\right) \mathmr{and} y + 3 = x - 2$

The equation of the line in standard form $\left(a x + b y = c\right)$ is y= x-5 or -x +y = -5 {Ans]