Question

Consider the straight lines x+2y+4=0x+2y+4=0 and 4x+2y−1=04x+2y−1=0. The line 6x+6y+7=06x+6y+7=0 is

 

a) bisector of the angle containing origin

 

b)bisector of acute angle

 

c) bisector of obtuse angle

 

d) bisector of angle not containing origin

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Answer 1

Step-by-step explanation:

Given that:

Consider the straight lines x+2y+4=0 and 4x+2y−1=0. The line 6x+6y+7=0 is

a) bisector of the angle containing origin

b)bisector of acute angle

c) bisector of obtuse angle

d) bisector of angle not containing origin

To find:Out of these four options which one is correct.

Solution:

It can be easily solved by plotting the graph of these three lines

Put the value of x and find the value of y,find at least three point for each line.Then plot graph for these three lines.

Graph is attached:

Blue line for :x+2y+4=0

Red line for:4x+2y-1=0

Yellow line for:6x+6y+7=0

From the graph it is clear that 6x+6y+7=0 does not contain origin.

Theoretically: Put the value of origin(0,0) in the eq. 6x+6y+7=0

6(0)+6(0)+7

0+0+7

7≠0

Thus,the line 6x+6y+7=0, does not contain origin.

As when two lines meet at a single point,total four angles are formed out of which two are obtuse and two are acute(If lines are not perpendicular)

From graph,

it is clear that 6x+6y+7=0 is bisector of acute angle.

Theoretically: Angle between two lines

x+2y+4=0,4x+2y-2=0

$tan\theta=\bigg|\frac{m_1-m_2}{1+m_1m_2}\bigg|\\\\m_1=\frac{-1}{2},m_2=-2\\\\tan\theta=\bigg|\frac{\frac{-1}{2}+2}{1+\frac{-1}{2}\times-2}\bigg|\\\\tan\theta=\frac{3}{4}\\\\\theta=tan^{-1}\frac{3}{4}=36.86°$

Angle between these two lines

x+2y+4=0,6x+6y+7=0

$tan\theta=\bigg|\frac{m_1-m_2}{1+m_1m_2}\bigg|\\\\m_1=\frac{-1}{2},m_2=-1\\\\tan\theta=\bigg|\frac{\frac{-1}{2}+1}{1+\frac{-1}{2}\times(-1)}\bigg|\\\\tan\theta=\frac{1}{3}\\\\\theta=tan^{-1}\frac{1}{3}=18.43°$

Thus,It is(6x+6y+7=0) bisector of acute angle not containing origin.

Hope it helps you.