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Question: Bisector of the angle between x+2y-1=0 and 2x+y+1=0 containing (2,1) is- A. x+y=0 B. x-y+1=0 C...

Bisector of the angle between x+2y-1=0 and 2x+y+1=0 containing (2,1) is-
A. x+y=0
B. x-y+1=0
C. x+y+2=0
D. x-y+2=0

Explanation

Solution

Hint: The equation of the angle bisector of two given lines is obtained by the formula-
a1x+b1y+c1a12+b12=±(a2x+b2y+c2a22+b22)\dfrac{{\mathrm a}_1\mathrm x+{\mathrm b}_1\mathrm y+{\mathrm c}_1}{\sqrt{{\mathrm a}_1^2+{\mathrm b}_1^2}}=\pm\left(\dfrac{{\mathrm a}_2\mathrm x+{\mathrm b}_2\mathrm y+{\mathrm c}_2}{\sqrt{{\mathrm a}_2^2+{\mathrm b}_2^2}}\right)

Complete step-by-step answer:
First, we will find the equation for both the angle bisectors and check which one satisfies (2,1).
Applying the formula for angle bisector-
x+2y112+22=±(2x+y+122+12)x+2y15=±(2x+y+15)x+2y1=2x+y+1  or  x+2y1=(2x+y+1)xy+2=0  or  3x+3y=0\dfrac{x+2y-1}{\sqrt{1^2+2^2}}=\pm\left(\dfrac{2x+y+1}{\sqrt{2^2+1^2}}\right)\\\\\dfrac{x+2y-1}{\sqrt5}=\pm\left(\dfrac{2x+y+1}{\sqrt5}\right)\\\\\mathrm x+2\mathrm y-1=2\mathrm x+\mathrm y+1\;\mathrm{or}\;\mathrm x+2\mathrm y-1=-\left(2\mathrm x+\mathrm y+1\right)\\\\\mathrm x-\mathrm y+2=0\;\mathrm{or}\;3\mathrm x+3\mathrm y=0


The green equations represent the two angle bisectors, the side which contains the point (2,1) will be the correct answer, which is the equation x - y = 2.
The correct option is D. x-y+2.

Note: It may be confusing to find the correct equations among the two answers. For this, either take the help of a diagram or substitute the point in both the equations and choose the correct answer according to the sign obtained at each equation. The best method is to draw a rough graph and check which bisector lies on the same side as the point