Question

Q1.(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

Step-by-step explanation:

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

$\sf{ {x}^{2} + 2xy - xy - 2 {y}^{2} = 0} </p><p>$

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

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

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:

Some common synonyms of chivy are badger, bait, heckle, hector, and hound. While all these words mean "to harass by efforts to break down," chivy suggests persecution by teasing or nagging

Step-by-step explanation:

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