Question

Find the equation of a line that is perpendicular to the line and contains the point

Answer 1

Given:

The equation of a first line is,

`y=(1)/(8)x+2`

The objective is to find the equation of a second line perpendicular to the first line through the point (-8,0).

Step-by-step explanation:

The general equation of straight line is,

`y=mx+b`

Here, m represents the slope of the line and b represents the y-intercept.

To find slope of first line:

By comparing the general equation with the first line,

`m_1=(1)/(8)`

To find slope of second line:

For perpendicular lines, the product of their slope values will be -1.

`\begin{gathered} m_1* m_2=-1 \\ (1)/(8)* m_2=-1 \\ m_2=-8_{} \end{gathered}`

Since the second line passes throught the point(-8,0), the equation of the line can be calculated using slope point formula of straight line.

`y-y_1=m_2(x-x_1)_{}`

To find equation of the perpendicular line:

Consider the given points as,

`(x_1,y_1)=(-8,0)`

On plugging the values in the above equation,

`\begin{gathered} y-0=-8(x-(-8)) \\ y=-8(x+8) \\ y=-8x-64 \end{gathered}`

Hence, the equation of the perpendicular line is y = -8x - 64.