Question

The straight lines 2x+3y-12 = 0, x+y-5 = 0 and x-2y+1 = 0 are concurrent.

• A. True
• B. False

Answer 1

The correct option is A

True

Three straight lines are concurrent means they meet at a point

We can find the intersection of any two lines first.We will check if the point of intersection is a point on the third line.

2x - 3y - 12 = 0 $\to$ (1)

x + y - 5 = 0 $\to$ (2)

$\Rightarrow$ y = 2

x = 3

so the point of intersection is (3,2)

Now, we have to check if (3,2) is a point on the line x - 2y + 1 = 0

3 - 4 + 1 = 0 $\Rightarrow$ (3,2) is a point on

x - 2y + 1 = 0

This means that the three lines are concurrent.

There is another way of checking concurrency.If $a_1$x+$b_1$y+$c_1$ = 0, $a_2$x+$b_2$y+$c_2$ = 0 and $a_3$x+$b_3$y+$c_3$ = 0 are concurrent,then

$\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|=0$

those who are familiar with determinants can use this method.In our example,

$\left|\begin{array}{ccc}2& 3& -12\\ 1& 1& -5\\ 1& -2& 1\end{array}\right|=0$

= 2(1-10) -3(1+5) -12(-2-1)

= -18 - 3 $\times$ 6 + 12 $\times$ 3

= 0

If any of those two lines are parallel, we will get the determinant value as zero.But it does not mean that three lines are concurrent.