Question

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

Answer 1

SOLUTION

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

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

Then the gradient of the other line is

$\begin{gathered} m_2=\frac{-1}{m_1} \\ m_2=\frac{-1}{5} \end{gathered}$

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

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

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

$y=-\frac{1}{5}x+4,,,,\text{ making y the subject of formula}$