If the straight lines ax + by + c = 0 and x cos(alpha) + y sin(alpha) = c enclose an angle pi/4 between them and meet the straight line x sin(alpha) - y cos(alpha) = 0 in the same point between them, then?

A) a^2 + b^2 = c^2
B) a^2 + b^2 = 2
C) a^2 + b^2 = 2c^2
D) a^2 + b^2 = 4

3 Answers
Dec 21, 2017

a^2+b^2=2

Explanation:

From the lines

L_1->ax+by+c=0
L_2->x cosalpha+ysinalpha +c=0

we have

a cos alpha+b sin alpha = sqrt(a^2+b^2) cos(pi/4) or

( a cos alpha+b sin alpha )^2= (a^2+b^2)/2

but from

{(x cosalpha+ysinalpha +c=0),(x sin alpha-y cos alpha=0):}

the intersection point gives us x=c cos alpha, y = c sin alpha

and after substituting into

ax+by+c=0 rArr a cosalpha + b sin alpha = -1

then

a^2+b^2=2

Dec 22, 2017

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Given lines are

L_1->ax + by + c = 0
L_2->x cos(alpha) + y sin(alpha) = c
L_3->x sin(alpha) - y cos(alpha) = 0

As per given condition above three given lines pass through a fixed point. L_1 subtends an angle pi/4 with L_2 and L_3 subtends an angle pi/2 with L_2 as the product of their slope =-1

Hence L_1 is the bisector of the angle between L_2 and L_3. Here two possible orientations of L_1 have been shown in the figure

Now equations of the angle bisectors between L_2 and L_3

(xsinalpha -ycosalpha)/sqrt(sin^2alpha+cos^2alpha)=pm(xcosalpha+ysinalpha-c)/sqrt(cos^2alpha+sin^2alpha)
=>xsinalpha -ycosalpha=pm(xcosalpha+ysinalpha-c)

Bisector -1
B_1->

x(sinalpha-cosalpha)-y(sinalpha+cosalpha)+c=0

Bisector 2

B_2->

x(sinalpha+cosalpha)+y(sinalpha-cosalpha)-c=0

Comparing L_1 and B_1 we get

a=sinalpha-cosalpha and b=-(sinalpha+cosalpha)

So a^2+b^2=2

Comparing L_1 and B_2 we get

a=-(sinalpha+cosalpha )and b=-(sinalpha-cosalpha)

So a^2+b^2=2

Dec 22, 2017

B) a^2+b^2=2

Explanation:

An alternative quick and dirty method to decide which of the given options is correct is to consider a particular example...

Let alpha = pi/4 and c = 3sqrt(2)

Then the third straight line is:

xsqrt(2)/2-ysqrt(2)/2 = 0

which simplifies to y=x

The second straight line is:

xsqrt(2)/2+ysqrt(2)/2 = 3sqrt(2)

which simplifies to x+y=6

These intersect at (3, 3)

The first straight line passes through this intersection point and is horizontal or vertical. That is: y=3 or x=3. Expressing these possibilities in the form ax+by+c=0 we get:

0x-sqrt(2)y+3sqrt(2) = 0

or:

-sqrt(2)x+0y+3sqrt(2) = 0

In either case a^2+b^2=2 and matches none of the other options.