Question

4. Consider the line:3y=7x-4a. Find the equation of the line that is parallel to the line above and passes through the point (9,-2)b. Find the equation of the line that is perpendicular to the line above and passes through thepoint (9,-2).

Answer 1

Given:

There is a line given as

`y=(3)/(7)x-4`

Required:

a)

Find equation of line that parallel which is passes through ( 9 , - 2 )

b)

Find equation of line that perpendicular which is passes through ( 9 , - 2 )

Step-by-step explanation:

Here take

`(x_1,y_1)=(9,-2)`

a)

The slope of parallel line is same

so the slope m is 3/7

now to find the line which is passes through ( 9 , - 2 )

`\begin{gathered} y-y_1=(3)/(7)(x-x_1) \\ \\ y+2=(3)/(7)(x-9) \\ \\ y=(3)/(7)x-(27)/(7)-2 \\ \\ y=(3)/(7)x-(27+14)/(7) \\ \\ y=(3)/(7)x-(41)/(7) \end{gathered}`

b)

Slope of perpendicular line is

`\begin{gathered} m^(\prime)=-(1)/(m) \\ \\ m^(\prime)=-(7)/(3) \end{gathered}`

now to find the line which is passes through ( 9 , - 2 )

`\begin{gathered} y-y_1=m^(\prime)(x-x_1) \\ y+2=-(7)/(3)(x-9) \\ \\ y=-(7)/(3)x+(7)/(3)*9-2 \\ \\ y=-(7)/(3)x+21-2 \\ \\ y=-(7)/(3)x+19 \end{gathered}`

Final answer:

a)

`y=(3)/(7)x-(41)/(7)`

b)

`y=-(7)/(3)x+19`