Question

Consider the line y=7x-7Find the equation of the line that is perpendicular to this line and passes through the point (-8,5) Find the equation of the line that is parallel to this line and passes through the point (-8,5)

Answer 1

Given:

The equation of a straight line is,

`y=7x-7`

The objective is to find,

a) The equation of perpendicular line passes throught the point (-8,5).

b) The equation of parallel line passes throught the point (-8,5).

Step-by-step explanation:

The general equation of straight line is,

`y=mx+c`

Here, m represents the slope of the straight line and c represents the y intercept.

a)

For perpendicular lines, the prouct of slope of two lines will be (-1).

By comparing the general equation and the given equation the slope value will be,

`m_1=7`

Now, the slope value of perpendicular line can be calculated as,

`\begin{gathered} m_1* m_2=-1 \\ 7* m_2=-1 \\ m_2=-(1)/(7) \end{gathered}`

Since, the perpendicular line passes through the point (-8,5), the equation of line can be calculated using point slope formula.

`\begin{gathered} y-y_1=m_2(x-x_1)_{} \\ y-5=-(1)/(7)(x-(-8)) \\ y-5=-(1)/(7)(x+8) \\ y-5=-(x)/(7)-(8)/(7) \\ y=-(x)/(7)-(8)/(7)+5 \\ y=-(x)/(7)-(8)/(7)+(35)/(7) \\ y=-(x)/(7)+(27)/(7) \end{gathered}`

Hence, the equation of perpendicular line is obtained.

b)

For paralle lines the slope value will be equal for both lines.

`m_1=m_3=7`

Since, the parallal line passes through the point (-8,5), the equation of line can be calculated using point slope formula.

`\begin{gathered} y-y_1=m_3(x-x_1) \\ y-5=7(x-(-8)) \\ y-5=7(x+8) \\ y-5=7x+56 \\ y=7x+56+5 \\ y=7x+61 \end{gathered}`

Hence, the equation of parallel line is obtained.