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Pavlova-9 [17]
3 years ago
6

A line passes through (-2,3), (2,5), and (6,k). Find k.

Mathematics
1 answer:
Semenov [28]3 years ago
4 0
We know the line runs through -2,3 and 2,5.

it also runs through 2, 5 and 6,k.

since all points are on the line, they're colinear, and therefore the line runs also through -2,3 and 6,k.

keeping in mind that line maintains a constant slope, therefore, the slope for -2,3 and 2,5, has to be the same slope as for 2,5 and 6,k.

what is the slope of -2,3  and 2,5 anyway?

\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&&(~{{ -2}} &,&{{ 3}}~) 
%  (c,d)
&&(~{{ 2}} &,&{{ 5}}~)
\end{array}
\\\\\\
% slope  = m
slope = {{ m}}\implies 
\cfrac{\stackrel{rise}{{{ y_2}}-{{ y_1}}}}{\stackrel{run}{{{ x_2}}-{{ x_1}}}}\implies \cfrac{5-3}{2-(-2)}\implies \cfrac{5-3}{2+2}\implies \cfrac{2}{4}\implies \cfrac{1}{2}

and since we know the slope of 2,5 and 6,k is the same, then,

\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&&(~{{ 2}} &,&{{ 5}}~) 
%  (c,d)
&&(~{{ 6}} &,&{{ k}}~)
\end{array}
\\\\\\
% slope  = m
slope = {{ m}}\implies 
\cfrac{\stackrel{rise}{{{ y_2}}-{{ y_1}}}}{\stackrel{run}{{{ x_2}}-{{ x_1}}}}\implies \cfrac{k-5}{6-2}\implies \cfrac{k-5}{4}=\stackrel{slope}{\cfrac{1}{2}}
\\\\\\
2k-10=4\implies 2k=14\implies k=\cfrac{14}{2}\implies k=7
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