Question

Write the equation for the line through (-1/2, -7/2) and (2,14) in slope intercept form

Answer 1

The slope-intercept form for the equation of a line is given by

`\begin{gathered} y=mx+c \\ \text{where m=slope and c=intercept} \end{gathered}`

Giving the point

`\begin{gathered} (-(1)/(2),-(7)/(2)) \\ \text{and} \\ (2,14) \end{gathered}`

Then

`\begin{gathered} x_1=-(1)/(2),y_1=-(7)/(2) \\ \text{and } \\ x_2=2,y_2=14 \end{gathered}`

We apply the two point-form for the equation of a line

`(y-y_1)/(x-x_1)=(y_2-y_1)/(x_2-x_1)`

Substituting the values into the formula, we obtain

`(y-(-(7)/(2)))/(x-(1)/(2))=(14-(-(7)/(2)))/(2-(-(1)/(2)))`

Simplify the equation above

`\begin{gathered} (y+(7)/(2))/(x-(1)/(2))=(14+(7)/(2))/(2+(1)/(2)) \\ \\ (y+(7)/(2))/(x-(1)/(2))=(28+7)/(2)\frac{.}{\text{.}}(4+1)/(2) \end{gathered}`

Then, change the division to multiplication and take the reciprocal of the last fraction

`\begin{gathered} (y+(7)/(2))/(x-(1)/(2))=(35)/(2)*(2)/(5) \\ (y+(7)/(2))/(x-(1)/(2))=7 \end{gathered}`

Multiply both parts of the equation by the denominator (x-1/2), we obtain

`\begin{gathered} y+(7)/(2)=7(x-(1)/(2)) \\ \text{expand the parenthesis} \\ y+(7)/(2)=7x-(7)/(2) \end{gathered}`

Then, make y the subject of the formula

`\begin{gathered} y+(7)/(2)=7x-(7)/(2) \\ \text{subtract 7/2 from both sides } \\ y=7x-(7)/(2)-(7)/(2) \\ y=7x-(14)/(2) \end{gathered}`

Hence the equation of the line becomes

`\begin{gathered} y=7x-7\ldots..\text{ in slope-intercept } \\ \text{slope}=7\text{ and intercept =-7} \end{gathered}`

Therefore the equation of the line in slope-intercept is

y=7x-7