Question

L1 is the line passing through (-2,14) and (6,10)

Answer 1

since you already know how to get the slopes and equation, let's do the equations of each of these two without much fuss.

then we'll have two equations, namely a system of equations of two variables, and we'll solve it by substitution.

first equation is

$\bf (\stackrel{x_1}{-2}~,~\stackrel{y_1}{14})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{10}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{10-14}{6-(-2)}\implies \cfrac{-4}{6+2}\implies -\cfrac{1}{2} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-14=-\cfrac{1}{2}[x-(-2)] \\\\\\ y-14=-\cfrac{1}{2}(x+2)\implies y=-\cfrac{1}{2}(x+2)+14$

second equation is

$\bf (\stackrel{x_1}{-1}~,~\stackrel{y_1}{-4})\qquad (\stackrel{x_2}{2}~,~\stackrel{y_2}{5}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{5-(-4)}{2-(-1)}\implies \cfrac{5+4}{2+1}\implies 3 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-(-4)=3[x-(-1)] \\\\\\ y+4=3(x+1)\implies y=3(x+1)-4$

now, let's recall, let's substitute "y" in the second equation,

$\bf \stackrel{\stackrel{y}{\downarrow }}{-\cfrac{1}{2}(x+2)+14}=3(x+1)-4\implies -\cfrac{x}{2}-1+14=3x+3-4 \\\\\\ -\cfrac{x}{2}+14=3x\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{2}}{-x+28=6x} \\\\\\ 28=7x\implies \cfrac{28}{7}=x\implies \blacktriangleright 4=x \blacktriangleleft \\\\[-0.35em] ~\dotfill$

$\bf \stackrel{\textit{now, let's substitute the found \underline{x} in the 2nd equation}}{y=3(4+1)-4\implies y=3(5)-4}\implies \blacktriangleright y= 11 \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill (4,11)~\hfill$