Question

lf the straight lines `ax + by + p` = 0 and `x cos alpha + y sin alpha = p` are inclined at an angle π/4 and concurrent with the straight line `x sin alpha - y cos alpha` = 0, then the value of `a^2 + b^2` is

Options

• 0

• 1

• 2

• None of these

Answer 1

2

Explanation:

Let `l_1 : ax + by + p` = 0

`l_2 : x cos alpha + y sin alpha = p`

`l_3 : x sin alpha - y cos alpha` = 0

⇒ It is given that lines `l_1` and `l_2` are inclined at an angle of `pi/4`,

`tan theta = |(m_1 - m_2)/(1 + m_1m_2)|`

Here, `theta = pi/4 , m_1 = (-a)/b, m_2 = (- cos alpha)/(sin alpha)`

`tan  pi/4 = 1 = |(- a/b + cosalpha/sinalpha)/(1 + a/b xx cosalpha/sinalpha)|`

`|(b cos alpha - a sin alpha)/(b sin alpha)| = |(b sin alpha + a cos alpha)/(b sin alpha)|`

∴ `|b cos alpha - a sin alpha| = |b sin alpha + a cos alpha|`

`b cos alpha - a sin alpha = +- (b sin alpha + a cos alpha)`

Squaring both the sides, we get

`b^2cos^2alpha + a^2sin^2alpha - 2ab sin alpha cos alpha`

= `b^2sin^2 alpha + a^2cos^2 alpha + 2ab sin alpha cos alpha` ......(i)

⇒ As it is given that, line `l_3` is concurrent with the line `l_1` and `l_2`. Then,

`l_2 : x cos alpha + y sin alpha = P` and `l_3 : x sin alpha - y cos alpha` = 0

 `x sin alpha = y cos alpha`

`y = (x sin alpha)/(cos alpha)`

Now substitute the value of y in line `l_2`,

`x cos alpha + (x sin alpha)/(cos alpha) xx sin alpha = P`

`x cos^2alpha + x sin^2alpha = P cos alpha`

`x (cos^2alpha + sin^2alpha) = P cos alpha`

∴ `x = P cos alpha`

Similarly, we get `y = P cos alpha`

We know that, line `l_3` is concurrent with `l_1` also.

∴ `ax + by + P = 0 ⇒ aP cos alpha + bP sin alpha + P` = 0

`a cos alpha + b sin alpha + 1` = 0

`a cos alpha + b sin alpha = -1`

Squaring the above equation, we get

`a^2cos^2alpha + b^2sin^2alpha - 2ab sin alpha cos alpha` = 1  .....(ii)

From equation (i) and (ii),

`a^2sin^2alpha + b^2cos^2 alpha - 2ab sin alpha cos alpha` = 1 ......(iii)

⇒ Now, adding equation (ii) and (iii), we get

`a^2sin^2alpha + b^2cos^2alpha + a^2cos^2alpha + b^2sin^2alpha` = 1 + 1

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

∴ `a^2 + b^2` = 2