Question

6Consider the line y=-=x+2Find the equation of the line that is perpendicular to this line and passes through the point (-6, 4).Find the equation of the line that is parallel to this line and passes through the point (-6, 4).Note that the ALEKS graphing calculator may be helpful in checking your answer.Hey

Answer 1

For a line to be perpendicular to another line, product of thier gradient must be -1.

if the gradient is given as

`\begin{gathered} m_{1\text{ }}andm_{2\text{ }},\text{ then} \\ m_1m_(2=-1) \end{gathered}`

Then given the line

`y=(6)/(7)x+2`

The gradient of the line is the coefficient of x using the expression

`\begin{gathered} y=mx+c \\ m=Gradient \end{gathered}`

Hence, we have

`m_1=(6)/(7)`

Then, using

`\begin{gathered} m_1m_2=-1 \\ m_2=(-1)/(m_1) \\ m_2=-1*(7)/(6)=-(7)/(6) \end{gathered}`

Given the point (-6,4), the line perpendicular will be having the equation

`\begin{gathered} y-y_1=m_2(x-x_1) \\ \text{where }y_1=4\text{ and x}_1=-6 \end{gathered}`

Then we obtain

`\begin{gathered} y-4=-(7)/(6)(x-(-6)) \\ y-4=-(7)/(6)x-7 \\ y=-(7)/(6)x-7+4 \\ y=-(7)/(6)x-3 \end{gathered}`

Therefore the equation perpendicular to this line is given as y=-7/6x - 3

`y=-(7)/(6)x-3`

Then the equation parallel to the same line will have the same gradient

`m_1=m_2=(6)/(7)`

Then the equation parallel passing through the point (-6,4) will be

`\begin{gathered} y-4=(6)/(7)(x-(-6)) \\ y-4=(6)/(7)(x+6) \\ y-4=(6)/(7)x+(36)/(7) \\ y=(6)/(7)x+(36)/(7)+4 \\ y=(6)/(7)x+(64)/(7) \end{gathered}`

The equation of the line that is parallel to this line and passes through the point (-6, 4) will be y=6/7x+64/7

`y=(6)/(7)x+(64)/(7)`