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* m_2=-1`

Substituting the value of m1, we will have

`\begin{gathered} 4* m_2=-1 \\ 4m_2=-1 \\ \text{divide both sides by 4} \\ (4m_2)/(4)=-(1)/(4) \\ m_2=-(1)/(4) \end{gathered}`

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

`m_2=(y-y_1)/(x-x_1)`

By substituting the values, we will have

`-(1)/(4)=(y-5)/(x-8)`

Cross multiply the equation above, we will have that

`\begin{gathered} -(1)/(4)=(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} \\ (4y)/(4)=-(x)/(4)+(28)/(4) \\ y=-(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