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Question

If the straight line through the point $$P(3,4)$$ makes an angle $$\dfrac{\pi}{6}$$ with the $$x-$$axis and meets the line $$12x+5y+10=0$$ at $$Q$$, find the length of $$PQ$$.


Solution

Any line through $$P(3,4)$$ making an angle  $$\dfrac{\pi}{6}$$ with $$x-$$axis is
$$\dfrac { x-3 }{ \cos { { 30 }^{ o } }  } =\dfrac { y-4 }{ \sin { { 30 }^{ o } }  } =r$$
Where $$r$$ represents the distance of any point on this line from the given point $$P(3,4)$$
Any point $$Q$$ on it is $${(r\sqrt {3}/2)+3,(r/2)+4}$$ and $$Q$$ lies on $$12x+5y+10=0$$
$$ \therefore\quad 12\left[ \left( \dfrac { r\surd 3 }{ 2 }  \right) +3 \right] +5\left( \dfrac { r }{ 2 } +4 \right) +10=0$$
$$\therefore\quad (12\sqrt {3}+5)r+132=0$$
$$\therefore\quad r=\dfrac { \left| -132 \right|  }{ 12\surd \left( 3 \right) +5 } =\dfrac { 132 }{ 12\surd \left( 3 \right) +5 } .$$

Mathematics

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