Which equation represents the line that passes through points (2, 0) and (6, -8)?

1 answer:

4 0

$\bf (\stackrel{x_1}{2}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{-8}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-8}-\stackrel{y1}{0}}}{\underset{run} {\underset{x_2}{6}-\underset{x_1}{2}}}\implies \cfrac{-8}{4}\implies -2 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{0}=\stackrel{m}{-2}(x-\stackrel{x_1}{2})\implies y=-2x+4$

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$\underline{\underline{\large\bf{Solution:-}}}\\$

$\longrightarrow$ Let us know about definition of altitude first. The altitude of a triangle is the perpendicular line segment drawn from the vertex to the opposite side of the triangle.

$\leadsto$ Median is the line segment from a vertex to the midpoint of the opposite side.

Let us Check all options one by one

CD is line segment which starts from vertex C but don't falls on opposite side AB thus it is not an altitude.❌

BA is line segment which starts from vertex B and falls perpendicularly on opposite sides AC and is thus an altitude.✔️

AD is line segment which starts from vertex A and falls perpendicularly on opposite side BC and is thus an altitude.✔️

AE is a line segment which starts from vertex A but doesn't falls perpendicularly on opposite side BC and is thus not an altitude.❌

AD falls on BC with D as mid point because BD = CD and is thus a median. ✔️

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