Question

In a $△ABC$, side $AB$ has the equation $2x+3y=29$ and the side $AC$ has the equation $x+2y=6$. If the mid - point of $BC$ is $(5,6)$ then the equation of $BC$ is

• A. $\text{x - y = - 1}$
• B. $\text{5x - 2y = 13}$
• C. $\text{21x + 31y = 291}$
• D. $\text{3x - 4y = - 9}$

Answer 1

The correct option is C $\text{21x + 31y = 291}$

In a triangle $ABC$, side $AB$ has the equation$2x+3y=29$ and the side $AC$ has the equation $x+2y=16$. If the mid point of $BC$ is $(5,6)$ the equation of $BC$ is

Given $AB=2x+3y=29$

$AC=x+2y=16$

$D(5,6)$

we know that $y=mx+c$

$2x+3y=29$

$3y=29-2x$

$y=\frac{29}{3}-\frac{2x}{3}$

then

$y=\frac{-2}{3}x+c\to \left(1\right)$

Substitute $(5,6)$ in equation $\left(1\right)$

$6=\frac{-2}{3}\left(5\right)+c$

$6=-\frac{10}{3}+c$

$18=-10+c$

$c=\frac{28}{3}$

Substitute $c$ in equation $\left(1\right)$

$y=\frac{-2}{3}x+\frac{28}{3}$

$3y=-2x+28$

$2x+3y=28---\left(2\right)$

$x+2y=16x---\left(3\right)$

Multiplying $2x\left(3\right)$

$2x+4y=32---\left(4\right)$

Solve $\left(2\right)$ & $\left(3\right)$

$\frac{2x+3=282x+4y=32---}{-y=-4}$

$y=4$

$y$ in equation $\left(3\right)$

$x+8=16$

$x=8$

then $y=mx+c$

$y=\frac{-1}{2}x+c$

$(5,6)\to 6=\frac{-1}{2}\left(5\right)+c\Rightarrow 12=5+2c$

$c=\frac{17}{2}$

$y=\frac{-1}{2}x+\frac{17}{2}\Rightarrow x+2y=17---\left(5\right)$

$2x+3y=29---\left(6\right)$

$\left(5\right)$ & $\left(6\right)$

Substitute $y$ in equation $\left(5\right)$

$x+2\left(5\right)=17$

$x=7$

$\frac{2x+2y=342x+3y=29---}{-y=-5y=5}$

$E(8,4)F(7,5)$

Slope $=\frac{5-4}{7-8}=\frac{+1}{-1}=-1$

$y=mx+c\Rightarrow y=x+c$

$(5,6)\to 6=-5+c=11$

$y=-x+11$

$x+y=11$ equation of $BC$