The straight lines $x+3 y-4=0, x+y-4=0$ and $3 x+y-4=0$
- form an isosceles triangle
- are concurrent
- form an equilateral triangle
- form a right angled isosceles triangle
Solution
$\begin{array}{ll} \text { So, } \quad & x+3 y-4=0 \\ & 3 x+y-4=0 \\ \Rightarrow \quad & 3 x+9 y-12=0 \end{array}$ $\begin{aligned} & \begin{array}{l}3 x+y-4=0 \\ -\quad-+\end{array} \\ & 8 y=8 \\ & y=1 \text { Now, } x=1\end{aligned}$ (ii) Now, for $B$ point $\begin{aligned} & x+3 y-4=0 \\ & x+y-4=0 \\ & \frac{-\quad+}{2 y=0} \\ & \Rightarrow \quad y=0, x=4 \\ & \end{aligned}$ (iii) for $C$ point $\begin{gathered} 3 x+y-4=0 \\ x+y-4=0 \\ -\quad-\quad+ \\ \hline 2 x=0 \\ x=0 \Rightarrow y=4 \end{gathered}$
Now, $\begin{aligned} & A B=\sqrt{9+1}=\sqrt{10} \\ & B C=\sqrt{32}=4 \sqrt{2} \\ & C A=\sqrt{10} \\ & \because \quad A B=C A=\sqrt{10} \\ & \end{aligned}$ $\therefore A B C$ an isosceles triangle.
Asked in: TEST SERIES MHT-CET Full Test 6