Question

Find an equation for the line that passes through the points (-5, -5) and (1, 3)

Answer 1

Solution
Let the lines be
(x1,y1)= (-5,-5)
(x2,y2)= (1,3)
Now
Using double point form
y-y1 =(y2-y1/x2-x1 )(x-x1)
or, y-(-5)= (3-(-5)/1-(-5)) (x-(-5))
or, y+5 = (3+5/1+5)(x+5)
or, y+5 = 8/6(x+5)
or, y+5 =4/3 (x+5)
or, 3(y+5) =4(x+5)
or, 3y+15=4x+20
or, 4x-3y+5= 0
Thus 4x-3y+5 =0 is the required equation,,

Answer 2

`(\stackrel{x_1}{-5}~,~\stackrel{y_1}{-5})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{3}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{3}-\stackrel{y1}{(-5)}}}{\underset{run} {\underset{x_2}{1}-\underset{x_1}{(-5)}}} \implies \cfrac{3 +5}{1 +5} \implies \cfrac{ 8 }{ 6 } \implies \cfrac{ 4 }{ 3 }`

`\begin{array}c \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-5)}=\stackrel{m}{ \cfrac{ 4 }{ 3 }}(x-\stackrel{x_1}{(-5)}) \implies y +5 = \cfrac{ 4 }{ 3 } ( x +5) \\\\\\ y+5=\cfrac{ 4 }{ 3 }x+\cfrac{ 20 }{ 3 }\implies y=\cfrac{ 4 }{ 3 }x+\cfrac{ 20 }{ 3 }-5\implies {\Large \begin{array}{llll} y=\cfrac{ 4 }{ 3 }x+\cfrac{5}{3} \end{array}}`