Question

What is the standard form equation of the line shown below?

Answer 1

Answer:

b

Step-by-step explanation:

You can determine the gradient from the two points (-2, 3) and (1, -3). The gradient of a straight line is:

$m=(y_2-y_1)/(x_2-x_1)$

$m=(-3-3)/(1-(-2))=-6/3=-2$

The equation for a straight line is:

$y=m\cdot{x}+c$

c is the y-intercept when x=0. We can substitute any point into the equation. Lets use point 1 and solve for c .

$3=-2\cdot{-2}+c$

$c=-1$

The equation is:

$y=-2\cdot{x}-1$

The answer is b

Answer 2

The point-slope form:

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

m - slope

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

We have the points (-2, 3) and (1, -3). Substitute:

$m=\dfrac{-3-3}{1-(-2)}=\dfrac{-6}{3}=-2\\\\y-(-3)=-2(x-1)\\\\y+3=-2(x-1)$

The standard form: Ax + By = C. Convert:

$y+3=-2(x-1)$      use distributive property

$y+3=-2x+2$         subtract 3 from both sides

$y=-2x-1$       add 2x to both sides

$\boxed{2x+y=-1}\to\boxed{c.}$