Question

5) through (-2, -3) and (0,1)Slope-Intercept:Standard Form: ​

Answer 1

For this case we have that by definition, the equation of the line of the slope-intersection form is given by:

$y = mx + b$

Where:

m: It is the slope of the line

b: It is the cut point with the y axis

We have the following points:

$(x_ {1}, y_ {1}): (0,1)\\(x_ {2}, y_ {2}): (-2, -3)$

We find the slope:$m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}} = \frac {-3-1} {- 2-0} = \frac {-4} {- 2} = 2$

Thus, the equation is of the form:

$y = 2x + b$

We substitute a point and find b:

$1 = 2 (0) + b\\b = 1$

Finally, we have:

$y = 2x + 1$

On the other hand, the equation in the standard form is given by:

$ax + by = c$

So, according to the slope-intersection equation we have:

$2x-y = -1$

Answer:

$y=2x+1\\2x-y=-1$

Answer 2

Answer:

Slope = 2

y-intercept = 1

x-intercept = -0.5

Standard Form ⇒ y - 2x = 1

Step-by-step explanation:

write the equation of the line through (-2 , -3) and (0,1)

The general form of the line is  y = mx + c

Where m is the slope and c is the y-intercept

The slope m = (y₂ - y₁)/(x₂ - x₁) = (1 - [-3])/(0 - [-2]) = 4/2 = 2

∴ y = 2x + c

By substitution with the point (0,1) to find c

1 = 2 *0 + c

c = 1

∴ y = 2x + 1

Or   y - 2x = 1 ⇒Standard Form

Also,

See the attached figure which represents the graph of the line y - 2x = 1