Question

Write an equation for a line perpendicular to y=2x+2 and passing through the point (6,-2)y =

Answer 1

The equation of the line is given as,

`y=2x+2`

According to the slope-intercept form, the equation of a line with slope 'm' and y-intercept 'c', is given by,

`y=mx+c`

Comparing with the given equation, the slope of the given line is,

`m=2`

Theorem: The product of two slopes of two perpendicular lines is -1 always.

Let the slope of the perpendicular line be (m'), and 'b' be the y-intercept. Then, it follows that,

`\begin{gathered} m^(\prime)\cdot m=-1 \\ m^(\prime)\cdot(2)=-1 \\ m^(\prime)=(-1)/(2) \end{gathered}`

The equation of this perpendicular line is given by,

`\begin{gathered} y=m^(\prime)x+b^(\prime) \\ y=((-1)/(2))x+b^(\prime) \end{gathered}`

Given that the perpendicular lines pass through the point (6,-2), so it must also satisfy its equation,

`\begin{gathered} -2=((-1)/(2))(6)+b^(\prime) \\ -2=-3+b^(\prime) \\ b^(\prime)=-2+3 \\ b^(\prime)=1 \end{gathered}`

Substitute this value back in the equation of the perpendicular line,

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

Thus, the equation of a line perpendicular to the given line and passing through the given point is obtained as,

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