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Question

Joint equation of pair of lines through(3,-2) and parallel to x2-4xy+3y2=0 is


A

x2+3y2-4xy-14x+24y+45=0

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B

x2+3y2+4xy-14x+24y+45=0

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C

x2+3y2+4xy-14x+24y-45=0

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D

x2+3y2+4xy-14x-24y-45=0

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Solution

The correct option is A

x2+3y2-4xy-14x+24y+45=0


Explanation for the correct option:

Step 1: Finding the values of m1 and m2

Given the equation is,x2-4xy+3y2=0

The given line is a combination of two lines, hence it can be written as,

x2-4xy+3y2=0

x2-3xy-xy-3y2=0x(x-3y)-y(x-3y)=0(x-3y)(x-y)=0x-3y=0......(i)x-y=0......(ii)

Let the slope of the equation (i) is m1.

m1=-1-3=13[m=-coeff.ofxcoeff.ofy]

Similarly, the slope of the equation (ii) is m2

m2=-1-1=1[m=-coeffofxcoeff.ofy]

Let the line parallel to the equation (i) be

x-3y+λ1=0.....(iii)

Let the line parallel to the equation (ii) be

x-y+λ2=0.....(iv)

These lines pass through a point 3,-2.

Step 2: Finding the values of λ1 and λ2

Substituting the value in equation (iii) and (iv) we get,

From equation (iii)

3-3-2+λ1=03+6+λ1=0λ1=-9

From equation (iv)

3--2+λ2=05+λ2=0λ2=-5

Therefore the equation (iii) will become

x-3y-9=0

Similarly, the equation (iv) will become

x-y-5=0

Step 3: Finding the equation of line

Now combined equation of (iii) and (iv) will be

x-3y-9x-y-5=0x2-xy-5x-3xy+3y2+15y-9x+9y+45=0x2+3y2-4xy-14x+24y+45=0

This is the joint equation of pair of lines passing through 3,-2 and parallel to x2-4xy+3y2=0.

Hence, option (A) is the correct answer.


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