Question

Write an equation of the line that passes through the given point and is perpendicular to the graph of the given equation.(4, 10); y=8x-1

Answer 1

Given:

`\begin{gathered} A\text{ point \lparen4,10\rparen} \\ Equation\text{ }y=8x-1 \end{gathered}`

Comparing the equation given to the equation of a line in slope-intercept form;

`\begin{gathered} y=mx+b \\ y=8x-1 \\ m=8\text{ is the slope} \\ b=-1\text{ is the y-intercept} \\ \\ Hence, \\ m_1=8 \end{gathered}`

For another line to pass through the point (4,10) perpendicular to the equation of the line given, then the slopes of both lines are negative reciprocals of each other.

This means;

`\begin{gathered} m_1m_2=-1 \\ m_2=-(1)/(m_1) \\ m_2=-(1)/(8) \end{gathered}`

The equation of a line through a point is given by;

`\begin{gathered} (y-y_1)/(x-x_1)=m_ \\ where; \\ x_1=4,y_1=10,m=-(1)/(8) \\ \\ Hence, \\ (y-10)/(x-4)=-(1)/(8) \\ Cross\text{ multiplying;} \\ 8(y-10)=-1(x-4) \\ 8y-80=-x+4 \\ 8y=-x+4+80 \\ 8y=-x+84 \\ \\ Dividing\text{ all through by 8;} \\ y=-(1)/(8)x+(84)/(8) \\ y=-(1)/(8)x+(21)/(2) \end{gathered}`

Therefore, the equation of the line is;

`y=-(1)/(8)x+(21)/(2)`