If \(x+2 y-3=0,3 x+4 y-7=0,2 x+3 y-4\) \(=0\) and \(4 x+5 y-6=0\) are the equations of four lines, then
If \(x+2 y-3=0,3 x+4 y-7=0,2 x+3 y-4\) \(=0\) and \(4 x+5 y-6=0\) are the equations of four lines, then
they are the sides of a square
they are all concurrent lines
they are the sides of a parallelogram
not all of them are concurrent
Solution
Given lines are
\(\begin{array}{r}
x+2 y-3=0 \quad \ldots (i) \\
3 x+4 y-7=0 \quad \ldots (ii) \\
2 x+3 y-4=0 \quad \ldots (iii) \\
4 x+5 y-6=0 \quad \ldots (iv)
\end{array}\)
On solving Eqs. (i) and (ii), we get point of intersection
\(\text{i.e. } \quad P=(1,1)\)
Here, \(P(1,1)\) is not satisfy Eqs. (iii) and (iv)
\(\therefore\) All the given lines are not concurrent.
\(\therefore\) Hence, answer is (d).