Question

(-5,-7) and (3,1) write equation in point slope form, intercept form, and standard form

Answer 1

Answer:

Please check the explanation

Step-by-step explanation:

Given the points

• (-5,-7)

• (3,1)

Determining the slope between the points:

$\mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}$

$\left(x_1,\:y_1\right)=\left(-5,\:-7\right),\:\left(x_2,\:y_2\right)=\left(3,\:1\right)$

$m=\frac{1-\left(-7\right)}{3-\left(-5\right)}$

$m=1$

Writing the equation in point-slope form

As the point-slope form of the line equation is defined by

$y-y_1=m\left(x-x_1\right)$

Putting the point $(-5,-7)$ and the slope $m = 1$ in the line equation

$y-y_1=m\left(x-x_1\right)$

$y-\left(-7\right)=1\left(x-\left(-5\right)\right)$

Hence, the equation in the point-slope form will be:

$y-\left(-7\right)=1\left(x-\left(-5\right)\right)$

Writing the equation in slope-intercept form

As the point-slope of the equation is

$y=mx+b$

where m is the slope and b is the y-intercept

Putting m = 1 and (3, 1) to determine the y-intercept

$y = mx+b$

$1 = 1 (3) + b$

$1 = 3 + b$

$b = -2$

so putting $b=-2$ and m = 1 in the slope-intercept form

$y = mx+b$

$y = (1)x + (-2)$

Therefore, the equation in slope-intercept form is:

$y = x -2$

Writing the equation in the standard form form

As we know that the equation in the standard form is

$Ax+By=C$

where x and y are variables and A, B and C are constants

As we already know the equation in slope-intercept form

$y = (1)x + (-2)$

so just simplify the equation to write in standard form

$y = (1)x + (-2)$

$y = x - 2$

$y - x = -2$