Question

Write an equation in slope-intercept form of a line passing through the given point andperpendicular to the given line.17. (4,2); x+y=118. (3,0);x-y=419. (-7,-3);2x+ 4y=8

Answer 1

The equation of a line in slope-intercept form is,

`y=mx+b`

Where,

m = slope

The given equation of the line is,

`x+y=1_{}`

Rearranging the given equation in a slope-intercept form in order to solve for the slope.

`\begin{gathered} x+y=1 \\ \therefore y=-x+1 \\ \text{where,} \\ m_1=-1 \end{gathered}`

We were told the equation of the line is perpendicular to the point given.

The rule for perpendicularism is,

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

Given

`\begin{gathered} (x_1,y_1)=(4,2) \\ m_2=1 \end{gathered}`

The formula to calculate the equation of the line in slope-intercept form perpendicular to the given point is,

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

Hence,

`\begin{gathered} y-2=1(x-4) \\ y=1(x-4)+2 \\ y=x-4+2 \\ \therefore y=x-2 \end{gathered}`

Therefore, the equation of the line in slope-intercept form is

`y=x-2`