Question

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

Then which of the following is true?

• A. they form a triangle
• B. they are concurrent
• C. one line bisects the angle between the other two
• D. two of them are parallel

Answer 1

Answer: (B), (C)

The correct options are
B they are concurrent
C one line bisects the angle between the other two

Given lines

$2x+11y-5=0$-------(1)

$24x+7y-20=0$-------(2)

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

On solving eq (1) and (2) we get

$x=\frac{37}{50}$ and $y=\frac{8}{25}$

Intersection Point $P(\frac{37}{50},\frac{8}{25})$

On solving eq (1) and (3) we get

$x=\frac{37}{50}$ and $y=\frac{8}{25}$

Intersection Point $Q(\frac{37}{50},\frac{8}{25})$

On solving eq (2) and (3) we get

$x=\frac{37}{50}$ and $y=\frac{8}{25}$

Intersection Point $R(\frac{37}{50},\frac{8}{25})$

So the intersection point of lines is same

$\therefore$ lines are concurrent

Eq of angle bisector of line (2) and (3)

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

$\frac{24x+7y-20}{\sqrt{625}}=\pm \frac{4x-3y-2}{\sqrt{25}}$

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

$24x+7y-20=\pm 5(4x-3y-2)$

$4x+22y-10=0$ or $44x-8y-30=0$

$2x+11y-5=0$ is same as eq (1)

hence line $2x+11y-5=0$ bisects the angle