please help me !!

Line q goes through points (-3, 3) and (-5, -3). At what point does line q cross the y-axis?

Answer:

(0, 12 )

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Calculate m using the slope formula

m = [tex]\frac{y_{2}-y_{1} }{x_{2}-x_{1} }[/tex]

with (x₁, y₁ ) = (- 3, 3) and (x₂, y₂ ) = (- 5, - 3)

m = [tex]\frac{-3-3}{-5+3}[/tex] = [tex]\frac{-6}{-2}[/tex] = 3 , then

y = 3x + c ← is the partial equation

To find c substitute either of the 2 points into the partial equation

Using (- 3, 3 ), then

3 = - 9 + c ⇒ c = 3 + 9 = 12 ← y- intercept

Thus the line crosses the y- axis at (0, 12 )

Answer:

It crosses the y-axis at the point where y = 12

- the point (0, 12).

Step-by-step explanation:

First we need to find the equation of the line q:

The slope of the line = (-3-3)/(-5-(-3))

= -6/-2

= 3.

Using the point-slope form of a line:

y - y1 = m(x - x1)         Take the point (-3, 3):

y - 3 = 3(x -(-3)

y - 3 = 3x + 9

y = 3x + 12.

Where q crosses the y-axis x = 0 so we have

y = 3(0) + 12

y = 12

The required point is (0, 12).

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