Question

Find the equation of the line that is perpendicular to y=5(x+1) and goes through (10, 2).

Answer 1

The equation of the line giving is

`y=5(x+1)`

The equation can be writting as

`y=5x+5`

From the equation of the line above the gradient of the line is

`\begin{gathered} m=5 \\ \text{where m=then gradient of the line by comparing it with } \\ y=mx+c \end{gathered}`

Since the line are perpendicular, the rule for perpendicularirt of two lines is

`\begin{gathered} m_1* m_2=-1 \\ \text{where m}_1=\text{ gradient of given line =5} \\ m_2=\text{ gradient of the other line } \end{gathered}`

Then the gradient of the other line is

`\begin{gathered} m_2=(-1)/(m_1) \\ m_2=(-1)/(5) \end{gathered}`

The equation of the line passing through the point (10,2 ) is given by

`\begin{gathered} y_{}-y_1=m(x-x_1) \\ \text{Where y}_1=2,x_1=10,\text{ m=-}(1)/(5) \\ \text{substitute the parameters, we obtain} \\ y-2=-(1)/(5)(x-10) \\ \end{gathered}`

Then by cross multiplying, we have

`\begin{gathered} 5(y-2)=-1(x-10)\ldots\text{ expand the parenthesis} \\ 5y-10=-x+10\ldots\ldots\text{ rearrange the expression } \\ 5y+x-10-10=0 \\ 5y+x-20=0 \end{gathered}`

Then the equation of the line that is perpendicular to y=5(x+1) and goes through (10, 2) is

`y=-(1)/(5)x+4,,,,\text{ making y the subject of formula}`