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Properties of Measures of Central Tendency

2 x 2 - 3 x ...

Question

$2 x^{2} - 3 x y + y^{2} + 5 x - 4 y + 3 = 0$ represent two straight lines of slope $m_{1}$ and $m_{2}$ Then choose correct options :-

harmonic mean of

m_{1}

and

m_{2}

\frac{4}{3}

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arithmetic

m_{1}

and

m_{2}

3

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point of intersection of these lines is

(0, - 1)

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None

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Solution

The correct option is A harmonic mean of $m_{1}$ and $m_{2}$ is $\frac{4}{3}$
$\begin{matrix} 2 x^{2} - 3 x y + y^{2} + 5 x - 4 y + 3 = 0 (2 x - y + c_{1}) (x - y + c_{2}) \equiv 2 x^{2} - 3 x y + y^{2} + 5 x - 4 y + 3 2 c_{2} + c_{1} = 5 c_{2} + c_{1} = 4 - - - - - - - c_{2} = 1 & c_{1} = 3 - - - - - - - - - - 2 x + 3 = y m_{1} = 2 x + 1 = y m_{2} = 1 H a r m o n i c m e a n o f m_{1} a n d m_{2} = \frac{2 \times 2 \times 1}{2 + 1} = \frac{4}{3} H e n c e, o p t i o n A i s t h e c o r r e c t a n s w e r . \end{matrix}$

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2x2−3xy+y2+5x−4y+3=0 represent two straight lines of slope m1 and m2 Then choose correct options :-

The correct option is A harmonic mean of m1 and m2 is 432x2−3xy+y2+5x−4y+3=0(2x−y+c1)(x−y+c2)≡2x2−3xy+y2+5x−4y+32c2+c1=5c2+c1=4−−−−−−−c2=1&c1=3−−−−−−−−−−2x+3=ym1=2x+1=ym2=1Harmonicmeanofm1andm2=2×2×12+1=43Hence,optionAisthecorrectanswer.

$2 x^{2} - 3 x y + y^{2} + 5 x - 4 y + 3 = 0$ represent two straight lines of slope $m_{1}$ and $m_{2}$ Then choose correct options :-