Question

The ends of the base of an isosceles right angle triangle are (1,2) , (3,4). The possible positions for the third vertex of the triangle are:

a) (2,3) , (3,2)
b) (1,4) , (3,2)
c) (1,4) , (-3,-2)
d) (3,2) , (4,3)

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Answer 1

Answer :-

( refer to the attachment )

Let the B be ( x , y )

Now, let's first calculate side BA and BC by distance formula :-

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$\sf Distance \: formula = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

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⇒ $\sf BA = \sqrt{(x - 1)^2 + (y-2)^2}$

⇒ $\sf BC = \sqrt{(x - 3)^2 + (y-4)^2}$

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Now, as it is a isoceles traingle,

⇒ BA = BC

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⇒ $\sf \sqrt{(x - 1)^2 + (y-2)^2} = \sqrt{(x - 3)^2 + (y-4)^2}$

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Squaring on both side :-

⇒ $\sf (\sqrt{(x - 1)^2 + (y-2)^2})^2 = (\sqrt{(x - 3)^2 + (y-4)^2})^2$

⇒ $\sf (x - 1)^2 + (y-2)^2 = (x-3)^2 + (y-4)^2$

⇒ $\sf \cancel{x^2} + 1 - 2x + \cancel{y^2} + 4 - 4y = \cancel{x^2} + 9 - 6x + \cancel{y^2} + 16 - 8y$

⇒ $\sf 1 - 2x + 4 - 4y = 9 - 6x + 16 - 8y$

⇒ $\sf 5 - 2x - 4y = 25 - 6x - 8y$

⇒ $\sf 6x - 2x + 8y - 4y = 25 - 5$

⇒ $\sf 4x + 4y = 20$

⇒ $\sf 4(x + y) = 20$

⇒ $\sf x + y = \frac{20}{4}$

⇒ $\sf x + y = 5$

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Checking from the option :-

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a) ( 2 , 3 ) , ( 3 , 2 )

⇒ 2 + 3 = 5

⇒ 3 + 2 = 5

Here, both are satisfying.

So, this option is correct.

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b) ( 1 , 4 ) , ( 3 , 2 )

⇒ 1 + 4 = 5

⇒ 3 + 2 = 5

Here, both are satisfying.

So, this option is also correct.

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c) ( 1 , 4 ) , ( - 3 , - 2)

⇒ 1 + 4 = 5

⇒ - 3 - 2 = -5

Here, both are not satisfying.

So, this option is incorrect.

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d) ( 3 , 2 ) , ( 4 , 3 )

⇒ 3 + 2 = 5

⇒ 4 + 3 = 7

Here, both are not satisfying.

So, this option is incorrect.

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Hence, option (a) and (b) are correct.

Answer 2

Step-by-step explanation:

Let the B be ( x , y )

$Now, let's first calculate side BA and BC by distance formula :-$

⠀

$\sf Distance \: formula = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

⠀

$⇒ \sf BA = \sqrt{(x - 1)^2 + (y-2)^2}$

$\sf BC = \sqrt{(x - 3)^2 + (y-4)^$

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Now, as it is a isoceles traingle,

⇒ BA = BC

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$⇒ \sf \sqrt{(x - 1)^2 + (y-2)^2} = \sqrt{(x - 3)$

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Squaring on both side :-

$⇒ \sf (\sqrt{(x - 1)^2 + (y-2)^2})^2 = (\sqrt{(x - 3)^2 + (y-4)^2})^2$

$⇒ \sf (x - 1)^2 + (y-2)^2 = (x-3)^2 + (y-4)^2(x−1)$

$⇒ \sf \cancel{x^2} + 1 - 2x + \cancel{y^2} + 4 - 4y = \cancel{x^2} + 9 - 6x + \cancel{y^2} + 16 - 8y$

$⇒ \sf 1 - 2x + 4 - 4y = 9 - 6x + 16 - 8y1−2x+4−4y=9−6x+16−8y</p><p></p><p>[tex]⇒ \sf 5 - 2x - 4y = 25 - 6x - 8y5−2x−4y=25−6x−8y$

$⇒ \sf 6x - 2x + 8y - 4y = 25 - 56x−2x+8y−4y=25−5$

$⇒ \sf 4x + 4y = 204x+4y=20$

$⇒ \sf 4(x + y) = 204(x+y)=20$

$⇒ \sf x + y = \frac{20}{4}x+y=$

$⇒ \sf x + y = 5x+y=5$

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Checking from the option :-

⠀

a) ( 2 , 3 ) , ( 3 , 2 )

⇒ 2 + 3 = 5

⇒ 3 + 2 = 5

Here, both are satisfying.

So, this option is correct.

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b) ( 1 , 4 ) , ( 3 , 2 )

⇒ 1 + 4 = 5

⇒ 3 + 2 = 5

Here, both are satisfying.

So, this option is also correct.

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c) ( 1 , 4 ) , ( - 3 , - 2)

⇒ 1 + 4 = 5

⇒ - 3 - 2 = -5

Here, both are not satisfying.

So, this option is incorrect.

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d) ( 3 , 2 ) , ( 4 , 3 )

⇒ 3 + 2 = 5

⇒ 4 + 3 = 7

Here, both are not satisfying.

So, this option is incorrect.

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Hence, option (a) and (b) are correct.