Question

Three straight lines $2x+11y-5=0$, $24x+7y-20=0$ and $4x-3y-2=0$

• A. form a triangle
• B. are only concurrent
• C. are concurrent with one line bisecting the angle between the other two
• D. None of the above

Answer 1

The correct option is C

are concurrent with one line bisecting the angle between the other two

Explanation for the correct answer:

Consider the equations of three straight lines

$2x+11y-5=0$ $...\left(i\right)$

$24x+7y-20=0$ $...\left(ii\right)$

$4x-3y-2=0$ $...\left(iii\right)$

Step 1: Prove that the lines are intersecting

As we can observe, that the slopes of all three lines are unequal (their coefficients of $x$ are different). Hence, the lines must be intersecting

Solve equations $\left(i\right)$ and $\left(iii\right)$ simultaneously to obtain the point of intersection

$4x+22y-10=0$

$-$ $4x-3y-2=0$

$\Rightarrow$ $25y=8$

$\Rightarrow$ $y=\frac{8}{25}$

Substitute this value of $y$ in $\left(i\right)$ we get,

$x=\frac{37}{50}$

Hence $\left(\frac{37}{50},\frac{8}{25}\right)$ is the point of intersection of lines $\left(i\right)$ and $\left(iii\right)$

Step 2: Prove that the lines are concurrent

Now substitute these values of $\left(x,y\right)$ in $\left(ii\right)$ we get,

$24\times \frac{37}{50}+7\times \frac{8}{25}-20=20-20=0$

Hence, point $\left(\frac{37}{50},\frac{8}{25}\right)$ satisfies the equation of line $\left(ii\right)$

Hence, the lines are concurrent.

Step 3: Use the formula for angle bisectors of a pair of lines

Angle bisectors of the lines $ax+by+c=0$ and $dx+ey+f=0$ are given as

$\frac{ax+by+c}{\sqrt{a^2+b^2}}=\pm \frac{dx+ey+f}{\sqrt{d^2+e^2}}$

Now, the angle bisectors of the lines $24x+7y-20=0$ and $4x-3y-2=0$ is given as

$\frac{24x+7y-20}{\sqrt{{24}^2+7^2}}=\pm \frac{4x-3y-2}{\sqrt{4^2+{\left(-3\right)}^2}}$

$\frac{24x+7y-20}{25}=\frac{4x-3y-2}{5}$ and $\frac{24x+7y-20}{25}=\frac{-4x+3y+2}{5}$

$\Rightarrow$ $24x+7y-20=20x-15y-10$ and $24x+7y-20=-20x+15y+10$

$\Rightarrow$ $4x+22y-10=0$ and $44x-8y-30=0$

$\Rightarrow$ $2x+11y-5=0$ and $22x-4y-15=0$

$2x+11y-5=0$ is the equation of line $\left(i\right)$

Hence, the lines are concurrent and one line is the angle bisector of the other two.

Hence, option (C) is the correct answer.