Question

Three lines

$L_1$ : x-y+6 = 0

$L_2$ _: 2x+y-3 = 0

$L_3$ : x-2y+m = 0 are given. Which of the following can be the value of m if $L_1$_, $L_2$ and $L_3$ form a triangle.

• A. 10
• B. 11
• C. 12
• D. 13

Answer 1

Answer: (A), (C), (D)

13

We can easily see that the three given lines are not parallel ( If the lines are parallel, their coefficients will be proportional). Three lines will form a triangle if they are not parallel and not concurrent.

We found that the given lines are not parallel. Now, if they are not concurrent also, we can say they form a triangle.

We will first find the condition for them to be concurrent and eliminate those values of m.[because we know how to find the value of m if they are concurrent]. We will find the point of intersection of first two lines and substitute that point in the third line to find m.

x-y+6 = 0

2x+y-3 = 0

3x+3 = 0

x = -1 and y = 5

(-1,5) lies on x-2y+m = 0

$\Rightarrow$ -1-10+m = 0

$\Rightarrow$ m = 11

Now we have to check if they are not parallel and concurrent if they will form a triangle. If two lines

$L_1$ and $L_2$ are not parallel, they will definitely meet at a point. Same thing is applicable for the pairs

($L_1$,$L_3$) and ($L_2$,$L_3$). So we will get three points of intersection. These three points won't be the same

because they are not concurrent. So they form a triangle.

So the three lines will be concurrent m = 11. For all the other value of m, the lines won't be concurrent and parallel.