Question

NEED HELP ITS A TEST!
Line n and point P are shown on the coordinate plane. What is the equation of the line parallel to line n that passes through point P?

A: y= 3x-6
B: y= 3x-6
C: y= 3x+12
D: y= -3x+3

Answer 1

Answer:

X = -2 - y/3

Step-by-step explanation:

First you need to know the current equation, for that:

Notice that when Y = 0, X = 2.

When Y = 6, X = 0

Therefore the equation is X = $2-\frac{y}{3}$

To move that to the left of the X axis (negative direction) you just have to reduce the X number:

$x=-2 - \frac{y}{3}$

Answer 2

Parallel lines have the same slope.

The required equation is: $\mathbf{y = -3x - 6}$

The points on line n are:

$\mathbf{(x,y) = (4,-6)(2,0)}$

So, the slope (m1) of line n is:

$\mathbf{m_1 =\frac{y_2 - y_1}{x_2 - x_1}}$

This gives

$\mathbf{m_1 =\frac{0--6}{2 - 4}}$

$\mathbf{m_1 =\frac{6}{-2 }}$

$\mathbf{m_1 = -3}$

If the line is parallel to line n, then it has the same slope (m2) as line n.

So, we have:

$\mathbf{m_2 = m_1 = -3}$

Point P is given as:

$\mathbf{P = (-3,3)}$

The equation is then calculated as:

$\mathbf{y = m_2(x - x_1) + y_1}$

This gives

$\mathbf{y = -3(x + 3) + 3}$

$\mathbf{y = -3x - 9 + 3}$

$\mathbf{y = -3x - 6}$

Hence, the required equation is: $\mathbf{y = -3x - 6}$

Read more about linear equations at:

https://brainly.com/question/11897796