Question

Given mn shown below with m (-6,1 ) and n (3,-5) what is the equation of the line that passes through point p (6,1) and is parallel to mn

Answer 1

First we need the slope of mn.  Using the slope formula, we find that it is -2/3.  So that's the "m" in y = mx + b.  Now we use the x and y from point p to find b, then rewrite the equation.  1 = -2/3(6) + b.  From this we find that b = 5. Therefore, the equation of the line that passes through p and is parallel to mn is y = -2/3x + 5

Answer 2

The equation of the line that passes through point p (6,1) and is parallel to MN is $\rm 3y+2x=15$.

Given that,

MN is shown below with m (-6,1 ) and n (3,-5)

We have to determine,

The equation of the line that passes through point p (6,1) and is parallel to MN.

According to the question,

To determine the equation of the line following all the steps given below.

MN is shown below with m (-6,1 ) and n (3,-5)

The slope of the line is given by,

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

Substitute the value in the formula;

$\rm m = \dfrac{y_2-y_1}{x_2-x_1}\\\\m = \dfrac{-5-1}{3-(-6)}\\\\m = \dfrac{-6}{9}\\\\m = \dfrac{-2}{3}$

The slope of the line is -2/3.

Therefore,

The equation of the line that passes through point p (6,1) and is parallel to MN.

$\rm y -y_1 = m(x-x_1)\\\\y-1=\dfrac{-2}{3} (x-6)\\\\3(y-1)= -2(x-6)\\\\3y-3=-2x+12\\\\3y+2x=12+3\\\\3y+2x=15$

Hence, The equation of the line that passes through point p (6,1) and is parallel to MN is $\rm 3y+2x=15$.

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