Question

write the equation of the line that goes through the point (8 5) and is perpendicular to the line y=4x+8

Answer 1

The general equation of a line is given below as

$\begin{gathered} y=mx+c \\ \text{where} \\ m=\text{Gradient} \\ c=\text{intercept on the y-a}\xi s \end{gathered}$

The equation of the line given in the question is

$y=4x+8$

By comparing coefficients, we will have the value of the gradient to be

$m_1=4$

The coordinates of the second line given are

$\begin{gathered} (8,5) \\ x_1=8,y_1=5 \end{gathered}$

Two lines are perpendicular when the products of their gradients is equal to -1, that is

$m_1\times m_2=-1$

Substituting the value of m1, we will have

$\begin{gathered} 4\times m_2=-1 \\ 4m_2=-1 \\ \text{divide both sides by 4} \\ \frac{4m_2}{4}=-\frac{1}{4} \\ m_2=-\frac{1}{4} \end{gathered}$

The formula to calculate the equation of a line when one point and the slope are given is

$m_2=\frac{y-y_1}{x-x_1}$

By substituting the values, we will have

$-\frac{1}{4}=\frac{y-5}{x-8}$

Cross multiply the equation above, we will have that

$\begin{gathered} -\frac{1}{4}=\frac{y-5}{x-8} \\ 4(y-5)=-1(x-8) \\ 4y-20=-x+8 \\ 4y=-x+8+20 \\ 4y=-x+28 \\ \text{divide all through by 4, we will have} \\ \frac{4y}{4}=-\frac{x}{4}+\frac{28}{4} \\ y=-\frac{x}{4}+7 \end{gathered}$

Hence,

The equation of the line perpendicular to y=4x+8 and passes through (8,5) =

y=-x/4 + 7