Question

(7) Show that one of the lines represented by x + xy - 2y = 0 passes through the point A(2,-1). Find the joint equation of to the lines lines passing through A(2,-1) and perpendicular represented by the equation x² + xy - 2y = 0.​

Answer 1

SOLUTION

TO DETERMINE

1. Show that one of the lines represented by x² + xy - 2y² = 0 goes through the point A(2,-1).

2. Find the joint equation of to the lines goes through A(2,-1) and perpendicular represented by the equation x² + xy - 2y² = 0.

EVALUATION

1. Here it is given that the lines represented by

x² + xy - 2y² = 0

Now x² + xy - 2y² = 0 gives

$\sf{ {x}^{2} + 2xy - xy - 2 {y}^{2} = 0}$

$\sf{ \implies \: x(x + 2y) - y(x + 2y)= 0}$

$\sf{ \implies \: (x + 2y) (x - y)= 0}$

So the pair of lines are x + 2y = 0 & x - y = 0

Since 2 + 2 × ( - 1 ) = 2 - 2 = 0

So the line x + 2y = 0 goes through A(2,-1)

But x - y = 0 is not satisfied by (2,-1)

So the line x - y = 0 does not goes through A(2,-1)

Hence one of the lines represented by x² + xy - 2y² = 0 goes through the point A(2,-1)

2. Here it is given that the lines represented by

x² + xy - 2y² = 0

Now x² + xy - 2y² = 0 gives the pair of lines are x + 2y = 0 & x - y = 0

Now the line perpendicular to x + 2y = 0 is 2x - y = c

Since 2x - y = c goes through the point A(2,-1)

So 4 + 1 = c

∴ c = 5

So the line is 2x - y = 5

Again the line perpendicular to x - y = 0 is x + y = k

Since x + y = k goes through the point A(2,-1)

So 2 - 1 = k

∴ k = 1

So the line is x + y = 1

Hence the joint equation of to the lines are

$\tt{(x + y - 1)(2x - y + 5) = 0}$

$\tt{ \implies \:2 {x}^{2} - {y}^{2} + xy - 7x - 4y + 5 = 0 }$

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Answer 2

Answer:

Step-by-step explanation:

1. Here it is given that the lines represented by

x² + xy - 2y² = 0

Now x² + xy - 2y² = 0 gives

So the pair of lines are x + 2y = 0 & x - y = 0

Since 2 + 2 × ( - 1 ) = 2 - 2 = 0

So the line x + 2y = 0 goes through A(2,-1)

But x - y = 0 is not satisfied by (2,-1)

So the line x - y = 0 does not goes through A(2,-1)

Hence one of the lines represented by x² + xy - 2y² = 0 goes through the point A(2,-1)

2. Here it is given that the lines represented by

x² + xy - 2y² = 0

Now x² + xy - 2y² = 0 gives the pair of lines are x + 2y = 0 & x - y = 0

Now the line perpendicular to x + 2y = 0 is 2x - y = c

Since 2x - y = c goes through the point A(2,-1)

So 4 + 1 = c

∴ c = 5

So the line is 2x - y = 5

Again the line perpendicular to x - y = 0 is x + y = k

Since x + y = k goes through the point A(2,-1)

So 2 - 1 = k

∴ k = 1

So the line is x + y = 1

Hence the joint equation of to the lines are