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➡Class 10th

➡Co-Ordinate Geometry
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Answer 1

$\huge{\pink{\underline{Answer \: 1}}}$

A(1,2)

B(4,3)

C(6,6)

D(3,5)

AB

$\sqrt{(4-1)^{2}+(3-2)^{2}}\\= \sqrt{3^{2} + (1)^{2}}\\=\sqrt{9+1}\\=\sqrt{10}$

BC

$\sqrt{(6-4)^{2}+(6-3)^{2}}\\= \sqrt{2^{2} + (3)^{2}}\\=\sqrt{4+9}\\=\sqrt{13}$

CD

$\sqrt{(3-6)^{2}+(5-6)^{2}}\\= \sqrt{(-3)^{2} + (-1)^{2}}\\=\sqrt{9+1}\\=\sqrt{10}$

DA

$\sqrt{(3-1)^{2}+(5-2)^{2}}\\= \sqrt{(2)^{2} + (3)^{2}}\\=\sqrt{4+9}\\=\sqrt{13}$

Checking diagonal's

AC

$\sqrt{(6-1)^{2}+(6-2)^{2}}\\= \sqrt{5^{2} + (4)^{2}}\\=\sqrt{25+16}\\=\sqrt{41}$

BD

$\sqrt{(3-4)^{2}+(5-3)^{2}}\\= \sqrt{(-1)^{2} + (2)^{2}}\\=\sqrt{1+3}\\=\sqrt{4}$

that's mean

The given is not a rectangle because the diagonal's of rectangle are always equal.

$\huge{\pink{\underline{Answer \: 2}}}$

A(0,0)

B(3,√3)

C(x,y)

$AB=CA=BC$

AB = CA

$\sqrt{(3-0)^{2}+(\sqrt3-0)^{2}} = \sqrt{(x-0)^{2} + (y-0)^{2}}\\(9+3)= x^{2} + y^{2}$

CA = BC

A(0,0) ; B(3,√3) ; C(x,y)

$\sqrt{(x-0)^{2} + (y-0)^{2}}= \sqrt{(x-3)^{2}+(y-\sqrt 3)^{2}}\\x^{2}+y^{2}=x^{2}+9-6x+y^{2}+3-2\sqrt y\\0=-6x+9+3-2\sqrt3 y\\-12= -6x - 2\sqrt3y \\6=(3x+\sqrt3y)\\2\sqrt3 = y + \sqrt3 x \\ y = 2\sqrt3 -\sqrt3x \\After \:solution\: we\: get\\y =\sqrt3(2-x)$

For some steps please refer attachment.

$x^{2} + (\sqrt3(2-x))^{2}=12\\ x^{2} + 3(4+x^{2}-4x) =12\\ x^2+\cancel{12}+3x^{2}-12x=\cancel{12}\\4x^{2}=12\\x = 3$

$y =\sqrt3 (2-x)\\y=\sqrt3 (2-3)\\ y = \sqrt3(-1) \\ y=-\sqrt3$