Line `a x+b y+p=0` makes angle `pi/4` with `xcosalpha+ysinalpha=p ,p in R^+` . If these lines and the line `xsinalpha-ycosalpha=0` are concurrent, then (a)`a^2+b^2=1` (b) `a^2+b^2=2` (c)`2(a^2+b^2)=1` (d) none of these

Options

A. `a^(2) +b^(2) = 1`
B. `a^(2) +b^(2) = 2`
C. `2(a^(2) +b^(2)) = 1`
D. none of these

Answer

B

Explanation

`"Lines "x"cos" alpha + y"sin"alpha =p " and "x"sin"alpha-y"cos"alpha=0` are mutually perpendicular. Thus, ax+by+p=0 will be equally inclined to these lines and would be the angle bisesctor of these lines. Now, the equations of angle bisectors are
`x"sin" alpha - y"cos"alpha =+-(x"cos"alpha+y"sin"alpha-p`
`"or " x("cos" alpha - "sin"alpha)+ y("sin"alpha+"cos"alpha)=p`
`"or " x("sin" alpha + "cos"alpha)- y("cos"alpha-"sin"alpha)=p`
Comparing these lines with ax+by+p=0, we get
`(a)/("cos" alpha-"sin"alpha) = (b)/("sin" alpha+"cos"alpha)= 1`
`rArr a^(2) + b^(2) = 2`
`(a)/("sin" alpha+"cos"alpha) = (b)/("sin" alpha-"cos"alpha)= 1`
`rArr a^(2) + b^(2) = 2`