Question

Write an equation of the line that pass through the given point and is (a) parallel and (b) perpendicular to the given line.

Answer 1

The answer is ( 2 , 2 )

Answer 2

Answer:

Step-by-step explanation:

Here we need to write the equations that passes through the point (4 , 3) and is parallel to the given line and perpendicular to it. So lets get started,

a)

First we calculate the slope of the given line by using any two points on the given line by the help of the formula:

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

we take the points (1 , 6) and (2 , 2) i chose these two you can choose any other two doesn't matter so now we calculate the slope,

$m=\frac{y_2-y_1}{x_2-x_1}\\\\m=\frac{2-6}{2-1}\\\\m=\frac{-4}{1}\\\\m=-4$

now since our slope is -4 and the line is parallel to the given line so the line that we need in part a would also have the same slope by the law parallel lines have same slope/gradient.

$m_1=m_2$

so the slope/gradient of our new line that we need which passes through the point (4,3) is -4

Now using the point slope form formula:

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

we have our point (4,3) which corresponds to $(x_1,y_1)$ and m which corresponds to -4 , so now

$y-y_1=m(x-x_1)\\y-3=-4(x-4)\\y-3=-4x+16\\y=-4x+19$

sow our new line in part a is

$y=-4x+19$

b)

In this part we need a line that is perpendicular to the given line which means we use the formula to calculate the slope of the perpendicular line which is:

$m_1m_2=-1$

where $m_1=-4$ the slope of our given line so the slope of the perpendicular line is,

$m_1m_2=-1\\-4m_2=-1\\\\m_2=\frac{1}{4}$

so using the point (4,3) and the new slope 1/4 we find the equation of the perpendicular line again by using the point-slope form formula:

$y-y_1=m(x-x_1)\\y-3=\frac{1}{4}(x-4)\\\\4y-12=x-4 \\4y=x+8\\\\y=\frac{x+8}{4}\\\\y=\frac{x}{4}+2$

I have attached of the graphs as well you can check it out for better understanding.