IIT JEE 1983 Maths - True or False Q5

Compute the intersection of the lines $x + 2y - 10 = 0$ and $2x + y + 5 = 0$

Multiply equation $1$ by $-2$ then add to equation $2$ .

$-2x - 4y + 20 = 0$

$+$

$2x + y + 5 = 0$

$\text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}$

$-3y + 25 = 0$

$y =$ $\frac{25}{3}$

solve for $x$ ,

$x=-2$ $(\frac{25}{3}) + 10 =$ $\frac{-20}{3}$

Plug in the values of $x$ and $y$ to $5x + 4y = 0$ .

$5($ $\frac{-20}{3}) + 4($ $\frac{25}{3})=0$

$0 = 0$

$\boxed{TRUE}$

If these lines are passing through a single point , then these three lines are concurrent . if they are concurrent then

$\left| \begin{matrix} a_1 & b_1 & c1\\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{matrix} \right| = 0$

$\left| \begin{matrix} 5 & 4 & 0 \\ 1 & 2 & -10 \\ 2 & 1 & 5\end{matrix} \right| = 5(10+10)-4(5+20) = 0$

The lines are concurrent , hence $5x+4y=0$ passes through the intersection of $x+2y-10=0$ and $2x+y+5 =0$