How do you find slope, point slope, slope intercept, standard form, domain and range of a line for Line D (3, -7) (9,-3)?

Jan 24, 2018

Slope is $\frac{2}{3}$ , point slope form is $y + 7 = \frac{2}{3} \left(x - 3\right)$,
slope-intercept form is $y = \frac{2}{3} x - 9.$, standard form is
$2 x - 3 y = 27$.Domain: $x \in \mathbb{R} \mathmr{and} \left(- \infty , \infty\right)$. Range: $y \in \mathbb{R} \mathmr{and} \left(- \infty , \infty\right)$

Explanation:

The slope of the line passing through $\left(3 , - 7\right) \mathmr{and} \left(9 , - 3\right)$ is

$m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{- 3 + 7}{9 - 3} = \frac{4}{6} = \frac{2}{3}$

Point slope form of the line is $y - {y}_{1} = m \left(x - {x}_{1}\right)$ or

$y + 7 = \frac{2}{3} \left(x - 3\right)$

Let the equation of the line in slope-intercept form be $y = m x + c$

or $y = \frac{2}{3} x + c$ . The point (3,-7) will satisfy the equation . So,

$- 7 = \frac{2}{3} \cdot 3 + c \mathmr{and} c = - 7 - 2 = - 9$. Hence the equation of

the line in slope-intercept form is $y = \frac{2}{3} x - 9.$

The equation of the line in standard form is $3 y = 2 x - 27$ or

$2 x - 3 y = 27$.

Domain of the line $y = \frac{2}{3} x - 9.$ is any real value of $x$

Domain: $x \in \mathbb{R} \mathmr{and} \left(- \infty , \infty\right)$

Range of the line $y = \frac{2}{3} x - 9.$ is any real value of $y$

Range: $y \in \mathbb{R} \mathmr{and} \left(- \infty , \infty\right)$

graph{2/3x-9 [-40, 40, -20, 20]}