MCQ
રેખા $3x + y = \lambda \,\left( {\lambda \ne 0} \right)$ પર ઉગમબિંદુથી બનાવેલ લંબ $P$ છે. જો રેખા $x-$ અક્ષને બિંદુ $A$ અને $y-$ અક્ષને બિંદુ $B$ આગળ છેદે છે તો $BP : PA$ ની કિમત મેળવો.
- A$9: 1$
- B$1 : 3$
- C$1 : 9$
- D$3 : 1$
$\frac{{x - 0}}{3} = \frac{{y - 0}}{1} = r$
For foot of perpendicular
$r = \frac{{ - \left( {\left( {3 \times 0} \right) + \left( {1 \times 0} \right) - \lambda } \right)}}{{{3^2} + {1^2}}} = \frac{\lambda }{{10}}$
So, food of perpendicular $P = \left( {\frac{{3\lambda }}{{10}},\frac{\lambda }{{10}}} \right)$
Given the line meets $X$-asix at $A = \left( {\frac{\lambda }{3},0} \right)$
and meets $Y$-axis at $B = \left( {o,\lambda } \right)$
So,
$BP = \sqrt {{{\left( {\frac{{3\lambda }}{{10}}} \right)}^2} + {{\left( {\frac{\lambda }{{10}} - \lambda } \right)}^2}} \Rightarrow BP = \sqrt {\frac{{9{\lambda ^2}}}{{100}} + \frac{{81{\lambda ^2}}}{{100}}} $
$ \Rightarrow BP = \sqrt {\frac{{90{\lambda ^2}}}{{100}}} $
Now,$PA = \sqrt {{{\left( {\frac{\lambda }{3} - \frac{{3\lambda }}{{10}}} \right)}^2} + {{\left( {0 + \frac{\lambda }{{10}}} \right)}^2}} $
$PA = \sqrt {\frac{{{\lambda ^2}}}{{900}} + \frac{{{\lambda ^2}}}{{100}}} \Rightarrow PA = \sqrt {\frac{{10{\lambda ^2}}}{{900}}} $
Therefore $BP:PA = 3:1$
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