Mathematics, 03.07.2019 04:00 dmaxbexkham
L1 is the line passing through (-2,14) and (6,10) l2 is the line passing through (-1,-4) and (2,5) find the coordinates of the point where 2 lines intersect
Answers: 1
![\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](/tpl/images/0045/0144/a5a2c.png)
![\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](/tpl/images/0045/0144/dfe5a.png)
![\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](/tpl/images/0045/0144/34f3d.png)
![\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](/tpl/images/0045/0144/63392.png)
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L1 is the line passing through (-2,14) and (6,10) l2 is the line passing through (-1,-4) and (2,5) f...
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