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Given A(5,-8) and B(-6.2), find the point on the segment AB that is three-fourths of the way from A to B.

I know the answer is (-3.25,-.5) I just do not know how to get there.


Sagot :

[tex]\textit{internal division of a segment using a fraction}\\\\ A(\stackrel{x_1}{5}~,~\stackrel{y_1}{-8})\qquad B(\stackrel{x_2}{-6}~,~\stackrel{y_2}{2})~\hspace{8em} \frac{3}{4}\textit{ of the way from A to B} \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_2}{-6}-\stackrel{x_1}{5}~~,~~ \stackrel{y_2}{2}-\stackrel{y_1}{(-8)})\qquad \implies \qquad \stackrel{\stackrel{\textit{component form of}}{\textit{segment AB}}}{\left( -11~~,~~ 10 \right)} \\\\[-0.35em] ~\dotfill[/tex]

[tex]\left( \stackrel{x_1}{5}~~+~~\frac{3}{4}(-11)~~,~~\stackrel{y_1}{-8}~~+~~\frac{3}{4}(10) \right)\implies \left(5-\cfrac{33}{4}~~,~~ -8+\cfrac{15}{2} \right) \\\\\\ \left(\cfrac{20-33}{4}~~,~~\cfrac{-16+15}{2} \right)\implies \left(- \cfrac{13}{4}~~,~~-\cfrac{1}{2} \right)[/tex]