Question

The equation of the perpendiculars drawn from the origin to the lines represented by the equation \[2{{x}^{2}}-10xy+12{{y}^{2}}+5x-16y-3=0~\] is
A. \[6{{x}^{2}}+5xy+{{y}^{2}}=0\] B
B. \[6{{y}^{2}}+5xy+{{x}^{2}}=0\]
C. \[6{{x}^{2}}-5xy+{{y}^{2}}=0\] D
D. None of these

Answer 1

Hint: Comparing the given equation with the general equation of straight line we can easily determine the value of the coefficients. Here we have to use the condition for perpendicular straight line passes through the origin.



Formula Used:$y=mx+c$
$a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$



Complete step by step solution:The given equation is \[2{{x}^{2}}-10xy+12{{y}^{2}}+5x-16y-3=0~\]
The general equation of curve is given as $a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$.
Comparing the given equation with the equation of curve we can get the following values.
\[\text{a}=2,\text{b}=12,\text{g}=\dfrac{5}{2},\text{f}=-8,c=-3,h=-5\]
Thus the product of the slopes is ${{m}_{1}}{{m}_{2}}=\dfrac{a}{b}$.
Putting the value of $a,b$ we get-
$ {{m}_{1}}{{m}_{2}}=\dfrac{2}{12} \\$
$ {{m}_{1}}{{m}_{2}}=\dfrac{1}{6} \\ $
Thus the value of the sum of the slopes is ${{m}_{1}}+{{m}_{2}}=\dfrac{-2h}{b}$.
Putting the values of $b,h$ we get-

  $ {{m}_{1}}+{{m}_{2}}=\dfrac{-2\times \left( -5 \right)}{12} \\ $
 ${{m}_{1}}+{{m}_{2}}=\dfrac{10}{12} \\ $
 $ {{m}_{1}}+{{m}_{2}}=\dfrac{5}{6} \\ $

The value of ${{m}_{1}}$ in terms of ${{m}_{2}}$ can be given as-
${{m}_{1}}=-{{m}_{2}}+\dfrac{5}{6}$
Putting this value in the equation ${{m}_{1}}{{m}_{2}}=\dfrac{1}{6}$ we get-

  $ {{m}_{1}}\left( \dfrac{5}{6}-{{m}_{1}} \right)=\dfrac{1}{6} \\ $
 $ -{{m}_{1}}^{2}+\dfrac{5}{6}{{m}_{1}}-\dfrac{1}{6}=0 \\ $
 $ 6{{m}_{1}}^{2}-5{{m}_{1}}+1=0 \\ $
 $ 6{{m}_{1}}^{2}-3{{m}_{1}}-2{{m}_{1}}+1=0 \\$
 $ 3{{m}_{1}}\left( 2{{m}_{1}}-1 \right)-1\left( 2{{m}_{1}}-1 \right)=0 \\ $
 $ \left( 2{{m}_{1}}-1 \right)\left( 3{{m}_{1}}-1 \right)=0 \\ $
 $ {{m}_{1}}=\dfrac{1}{3},{{m}_{1}}=\dfrac{1}{2} \\ $

Now we know that the slopes of the perpendicular lines are negative reciprocals of the slopes. Thus the slopes of perpendicular lines are $-2,-3$ .
Thus the equations of the separate lines are
  $ y+2x=0 \\$
 $ y+3x=0 \\ $

Thus the equation of the pair of straight lines is given as-
$ (y+2x)(y+3x)=0 \\ $
 ${{y}^{2}}+2xy+3xy+6{{x}^{2}}=0 \\$
 $ {{y}^{2}}+5xy+6{{x}^{2}}=0 \\$




Option ‘A’ is correct



Note: The another method to solve this equation is to determine the sum of the coefficients $a,b$ . If the sum of $a,b$ is zero then the straight lines are perpendicular to each other.