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Maths-

General

Easy

Question

Line $a x plus b y plus p equals 0$ makes angle $pi divided by 4$ with $x blank c o s blank alpha plus y blank s i n blank alpha equals p comma p space element of R to the power of straight prime$ If these lines and the line $x blank s i n blank alpha minus y blank c o s blank alpha equals 0$ are concurrent, then

$a^{2} + b^{2} = 1$
$a^{2} + b^{2} = 2$
2 $(a^{2} + b^{2}) = 1$
$a^{2} - b^{2} = 1$

The correct answer is: $a to the power of 2 end exponent plus b to the power of 2 end exponent equals 2$

We should find some relation between a and b using given conditions
Lines $x cos α + y sin α = p$ and $x sin α - y cos α = 0$ are mutually perpendicular. Thus $a x + b y + p = 0$ will be equally inclined to these lines and would be the angle bisector of these lines. Now equations of angle bisectors is
$x sin α - y cos α = \pm (x cos α + y sin α - p)$
$x (cos α - sin α) + y (cos α + sin α) = p$
or
$x (sin α + cos α) - y (cos α - sin α) = p$
Comparing these lines with $a x + b y + p = 0$ , we get
$cos α - sin α a = sin α + cos α b = - 1$
$a 2 + b 2 = 2$
or
$cos α + sin α a = - sin α + cos α b = - 1$
$a 2 + b 2 = 2$

Hence option 2 is correct

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