Question

(A) Find the slope of the line that passes through the given points.(B) Find the standard form of the equation of the line.(C) Find the slope-intercept form of the equation of the line.(5,3) and (11,8)

Answer 1

Given that a line passes through points below

$\begin{gathered} (x_1,y_1)\Rightarrow(5,3)_{} \\ (x_2,y_2)\Rightarrow(11,8) \end{gathered}$

A) To find the slope, m, of a line, the formula is

$m=\frac{y_2-y_1}{x_2-x_1}$

Substitute the coordinates into the formula above

$\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ m=\frac{8-3}{11-5}=\frac{5}{6} \end{gathered}$

Hence, the slope, m, is 5/6

B) To find the equation of a line, the formula is

$\begin{gathered} \frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1} \\ \frac{y-3}{x-5}=\frac{5}{6} \\ \text{Crossmultiply} \\ 6(y-3)=5(x-5) \\ 6y-18=5x-25 \\ 6y-5x=-25+18 \\ -5x+6y=-7 \end{gathered}$

The standard form of an equation of a straight line is

$Ax+By=C$

Hence, the equation of the line in standard form is -5x + 6y = -7

C) The slope-intercept form of the equation of a straight line is

$y=mx+b$

Make y the subject

$\begin{gathered} -5x+6y=-7 \\ 6y=5x-7 \\ \text{Divide both sides by 6} \\ \frac{6y}{6}=\frac{5x-7}{6} \\ y=\frac{5}{6}x-\frac{7}{6} \end{gathered}$

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

$y=\frac{5}{6}x-\frac{7}{6}$