Question

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) then the equation of BC is

• A. x y = 1
• B. 5x 2y = 13
• C. x + y = 11
• D. 3x 4y = 9

Answer 1

The correct option is C x + y = 11

$\begin{array}{c}\text{Solution -}\\ \begin{array}{cc}& AB:2x+3y=29-\left(1\right),BC=(5,6)\\ & AC:x+2y=16-\left(2\right)\end{array}\end{array}$

$\begin{array}{cc}\therefore DE& \Rightarrow y=mx+c\\ & \Rightarrow y=-\frac{2}{3}x+c\\ (5,6)& \text{mill satisfy the equation}\\ & 6=-\frac{2}{5}\left(5\right)+c\\ \Rightarrow & c=\frac{28}{3}\end{array}$

$\therefore$

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

$\Rightarrow 3y+2x=28$

Solve equation (2) and (3) we get

$y=4;x=8$

Similarly $DF\Rightarrow y=mx+C$

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

$(5,6)\to 6=-\frac{1}{2}5+C\Rightarrow C=\frac{17}{2}$

$\begin{array}{cc}\text{Similarly}\Delta F& \Rightarrow y=mx+c\\ & \Rightarrow y=-\frac{1}{2}x+C\\ (5,6)\to 6& =-\frac{1}{2}5+c\Rightarrow c=\frac{17}{2}\\ \text{hence}DF:x+2y=17-& \text{(4)}\\ \text{Solve equation}\left(1\right)\text{and}\left(4\right)\text{nu get}& \\ x& =7;y=5\end{array}$

$\begin{array}{c}\text{Equation of Ef where}E(8,4)\text{and}f(7,5)\\ \text{useing two point form}\\ \begin{array}{c}(y-4)=\left(\frac{5-4}{7-8}\right)(x-8)\\ y-4=8-x\end{array}\\ \Rightarrow x+y=12\end{array}$

$\begin{array}{c}\text{Now,}BC\text{is}11\text{to EF}\\ \therefore BC:x+y=K\\ \begin{array}{c}(5,6)\text{lies on}BC\\ \therefore 5+6=k\Rightarrow K=11\end{array}\\ \text{hence equation of}BC:x+y=11\\ \therefore \left(c\right)\text{is the correct option.}\end{array}$