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Ghella [55]
2 years ago
6

given abc with a(-4 -2), b(4,4), and c(18,-8). write the equation of the line that contains the altitude that passes through b i

n standard form

Mathematics
1 answer:
Tresset [83]2 years ago
4 0

check the picture below.

so red line of BD is perpendicular to AC, hmmmm let's firstly find the slope of AC, bearing in mind that perpendicular lines have <u>negative reciprocal</u> slopes.


\bf A(\stackrel{x_1}{-4}~,~\stackrel{y_1}{-2})\qquad C(\stackrel{x_2}{18}~,~\stackrel{y_2}{-8}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-8-(-2)}{18-(-4)}\implies \cfrac{-8+2}{18+4} \\\\\\ \cfrac{-6}{22}\implies -\cfrac{3}{11} \\\\[-0.35em] \rule{34em}{0.25pt}

\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{-\cfrac{3}{11}}\qquad \qquad \qquad \stackrel{reciprocal}{-\cfrac{11}{3}}\qquad \stackrel{negative~reciprocal}{\cfrac{11}{3}}}\impliedby \textit{BD's slope}


so we're really looking  for the equation of a line whose slope is 11/3 and runs through B(4,4).  Keeping in mind that

standard form for a linear equation means

• all coefficients must be integers, no fractions

• only the constant on the right-hand-side

• all variables on the left-hand-side, sorted

• "x" must not have a negative coefficient


\bf B(\stackrel{x_1}{4}~,~\stackrel{y_1}{4})~\hspace{10em} slope = m\implies \cfrac{11}{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=\cfrac{11}{3}(x-4)


\bf \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{3}\textit{ to do away with the denominators}}{3(y-4)=3\left( \cfrac{11}{3}(x-4) \right)\implies 3y-12=11(x-4)} \\\\\\ 3y-12=11x-44\implies 3y=11x-32 \\\\\\ -11x+3y=-32\implies \blacktriangleright 11x - 3y= 32\blacktriangleleft

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