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

Line `a x+b y+p=0`makes angle `pi/4`with `cosalpha+ycosalpha+ysinalpha

1.	Line `a x+b y+p=0`makes angle `pi/4`with `cosalpha+ycosalpha+ysinalpha=p ,p in R^+`. If these lines and the line `xsinalpha-ycosalpha=0`are concurrent, then`a^2+b^2=1`(b) `a^2+b^2=2``2(a^2+b^2)=1`(d) none of theseA. `a^(2) +b^(2) = 1`B. `a^(2) +b^(2) = 2`C. `2(a^(2) +b^(2)) = 1`D. none of these
Answer» Correct Answer - B `"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`

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

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