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1.Verify whether the points (1,5),(2,3) and (-2,-1)

2.Find the relation between x and y such that point (x,y) is equidistant from the points (7,1)and (3,5).

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Answer 1

$\huge \fbox \red{Answer}$

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1. Let the points (1,5), (2,3) and (-2,-1) be represented by A,B and C.

A(1,5), B(2,3), C(-2,-1)

$AB = \sqrt{ {(2 - 1)}^{2} + {(3 - 5)}^{2} }$

$= \sqrt{ {(1)}^{2} + {( - 2)}^{2} }$

$= \sqrt{1 + 4}$

$= \sqrt{5}$

$BC = \sqrt{ {( - 2 - 2)}^{2} + {( - 1 - 3)}^{2} }$

$= \sqrt{ {( - 4)}^{2} + {( - 4)}^{2} }$

$= \sqrt{16 + 16}$

$= \sqrt{32}$

$CA = \sqrt{ {( - 2 - 1)}^{2} + {( - 1 - 5)}^{2} }$

$= \sqrt{ { (- 3)}^{2} + { (- 6)}^{2} }$

$= \sqrt{9 + 36}$

$= \sqrt{45}$

As we can see that

AB + BC ≠ CA

Therefore, the point (1,5), (2,3) and (-2,-1) are not collinear.

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2. Let the P(x,y) be equidistant from A(7,1) and B(3,5).

P(x,y), A(7,1) and B(3,5)

Then,

PA = PB

PA² = PB²

↪(7 - x)² + (1 - y)² = (3 - x)² + (5 - y)²

↪x² + y² - 14x - 2y + 50 = x² + y² - 6x - 10y + 34

↪8x - 8y = 16

↪x - y = 2

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Answer 2

$\huge\orange\star~\huge \sf\green {\boxed{\sf{ \red{Ꭺ{ \pink{ɳ{ \blue{s{ \purple {ฬ{ \orange{є { \red{Ʀ\pink{~ :}}}}}}}}}}}}}}}$

Let the points (1,5), (2,3) and (-2,-1) be represented by A,B and C.

A(1,5), B(2,3), C(-2,-1)

$\tt {\red{ \underline{AB = \sqrt{ {(2 - 1)}^{2} + {(3 - 5)}^{2} } }}}$

$\: \: \: \: \: \implies{ \sf\sqrt{ {(1)}^{2} + {( - 2)}^{2} } }$

$\: \: \: \: \: \implies \sqrt{1 + 4}$

$\: \: \: \: \: \implies \sqrt{5}$

$\tt \red{ \underline{BC = \sqrt{ {( - 2 - 2)}^{2} + {( - 1 - 3)}^{2} }}}$

$\: \: \: \: \: \implies \sf{ \sqrt{ {( - 4)}^{2} + {( - 4)}^{2} }}$

$\: \: \: \: \: \implies {\sf \sqrt{16 + 16} }$

$\: \: \: \: \: \implies \sf \sqrt{32}$

${ \sf {\red{ \underline{ \: CA =\sqrt{ {( - 2 - 1)}^{2} + {( - 1 - 5)}^{2} } }}}}$

$\: \: \: \: \: \implies \sf\sqrt{ { (- 3)}^{2} + { (- 6)}^{2} }$

$\: \: \: \: \: \implies \sf\sqrt{9 + 36}$

$\: \: \: \: \: \implies \sf \sqrt{45}$

As we can see that

$\boxed{ \blue{\tt{AB + BC ≠ CA}}}$

Therefore, the point (1,5), (2,3) and (-2,-1) are not collinear.

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$\huge\orange\star~\huge \sf\green {\boxed{\sf{ \red{Ꭺ{ \pink{ɳ{ \blue{s{ \purple {ฬ{ \orange{є { \red{Ʀ\pink{~ :}}}}}}}}}}}}}}}$

Let the P(x,y) be equidistant from A(7,1) and B(3,5).

P(x,y), A(7,1) and B(3,5)

Then,

➞ PA = PB

➞ PA² = PB²

➞ (7 - x)² + (1 - y)² = (3 - x)² + (5 - y)²

➞ x² + y² - 14x - 2y + 50 = x² + y² - 6x - 10y + 34

➞ 8x - 8y = 16

$\large \boxed {\pink{ \tt{ x - y = 2}}}$