Question bank/JEE Main/Mathematics

If the straight line ax + by + p =0 and xcosα + ysinα = p
enclosed an angle of$\frac{\pi}{4}$and the line
xsinα - ycosα = 0 meets them at the same point, then a^2 +b^2 is
ax + by + p =0 rFkk xcosα + ysinα = p s.k$\frac{\pi}{4}$gS rFkk ljy js[kk
xsinα - ycosα = 0 mu s,d gh fcUnq ij feyrh gS] rc a^2
+b^2gS

A4B3C2D1

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[IMAGE] So from diagram
Perpendicular distance from origin to line ax + by + p = 0 is
$\frac{p}{\sqrt{2}}$then
$\left| \frac{p}{\sqrt{a^{2} + b^{2}}} \right| = \left| \frac{p}{\sqrt{2}} \right| \Rightarrow a^{2} + b^{2} = 2$
ewy fcUnq ls ljy js[kk ax + by + p = 0 ij Mkys x, yEc dh yEckbzZ
$\frac{p}{\sqrt{2}}$rc
$\left| \frac{p}{\sqrt{a^{2} + b^{2}}} \right| = \left| \frac{p}{\sqrt{2}} \right|$
$\Rightarrow a^{2} + b^{2} = 2$

Source: Grass-G / M-26 Mathematics Paper 1 15 Jan.docx