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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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Find acute as well as obtuse angular bisectors of the lines $x+3y+1=0$ & $2x+5y+2=0$.

$\begin{array}(a)\;\large\frac{x+3y+1}{\sqrt{10}}+\frac{2x+5y+2}{\sqrt{29}}=0,\large\frac{x+3y+1}{\sqrt{10}}-\frac{2x+5y+2}{\sqrt{29}}=0\\(b)\;\large\frac{x+2y+1}{\sqrt{13}}+\frac{2x+5y+2}{\sqrt{19}}=0,\large\frac{x+3y+1}{\sqrt{10}}-\frac{2x+5y+2}{\sqrt{23}}=0\\(c)\;\large\frac{x+3y-1}{\sqrt{10}}+\frac{2x+5y-2}{\sqrt{29}}=0,\large\frac{x+3y+1}{\sqrt{10}}-\frac{2x+5y-2}{\sqrt{29}}=0\\(d)\;\large\frac{x-3y-1}{\sqrt{10}}+\frac{2x+5y+2}{\sqrt{29}}=0,\large\frac{x-3y-1}{\sqrt{10}}-\frac{2x+5y+2}{\sqrt{29}}=0\end{array}$

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1 Answer

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  • The acute and obtuse angular bisectors of the line $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ are given by $\bigg|\large\frac{a_1x+b_1y+c_1}{\sqrt {a_1^2+b_1^2}}\bigg|$$=\pm\bigg|\large\frac{a_2x+b_2y+c_2}{\sqrt {a_2^2+b_2^2}}\bigg|$
Let us check the value of $a_1a_2+b_1b_2$
If $a_1a_2=b_1b_2 > 0$
Acute angle bisector $\rightarrow$ -ve sign
obtuse angle bisector $\rightarrow$ +ve sign
else if $a_1a_2=b_1b_2 < 0$
Acute angle bisector $\rightarrow$ +ve sign
obtuse angle bisector $\rightarrow$ -ve sign
$1\times 2+1\times 35$ which is greater than zero
Hence $\large\frac{\mid x+3y+1\mid}{\sqrt{10}}=\frac{\mid2x+5y+2\mid}{\sqrt{29}}$
Hence the acute angular bisector is of -ve sign
$\large\frac{x+3y+1}{\sqrt{10}}=\large\frac{-2x-5y-2}{\sqrt{29}}$
$\large\frac{x+3y+1}{\sqrt{10}}+\frac{2x+5y+2}{\sqrt{29}}=$$0$
and
Obtuse anguar bisector is
$\large\frac{x+3y+1}{\sqrt{10}}-\frac{2x+5y+2}{\sqrt{29}}$$=0$
Hence (a) is the correct answer.
answered Feb 4, 2014 by sreemathi.v
edited Mar 19, 2014 by rvidyagovindarajan_1
 

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