Question

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Answer 1

Answer:

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Answer 2

Solution:-

=> Let the ∆ABC and let the three side AB , AC and BC

Using Distance formula We get the length of side of triangle

Formula:-

$\boxed{\rm D = \sqrt{(x_2 - x_1) {}^{2} + (y_2 - y_1) {}^{2} } }$

Its given Coordinate

$\rm \implies \: A (5, - 2),B(6,4) \: \: and \: \: C(7, - 2)$

Now take a distance AB

$\rm \implies \: A(x_1 = 5,y_1 = - 2) \: and \: B(x_2 = 6,y_2 = 4)$

Put the value On Distance formula

$\rm \implies A B = \sqrt{(6 - 5) {}^{2} + (4 - ( - 2)) {}^{2}}$

$\rm \implies \:A B = \sqrt{ {1}^{2} + {6}^{2} }$

$\rm \implies \: A B = \sqrt{1 + 36} = \sqrt{37} sq \: unit$

Now Take a distance AC

$\rm \implies \: A(x_1 = 5,y_1 = - 2) \: and \: C(x_2 = 7,y_2 = - 2)$

$\rm \implies \: A C = \sqrt{(7 - 5) {}^{2} + \{ - 2 - ( - 2) \} {}^{2} }$

$\rm \implies \: AC \: = \sqrt{ {2}^{2} } = 2sq \: unit$

Now Take a distance BC

$\rm \implies \: B(x_1 = 6,y_1 = 4) \: and \: C(x_2 = 7,y_2 = - 2)$

$\rm \implies \:BC = \sqrt{(7 - 6)^{2} +( - 2 - 4) ^{2} }$

$\rm \implies \: BC = \sqrt{ {1}^{2} + {6}^{2} } = \sqrt{37} sq \: unit$

So AB and BC are equal and AC is not equal

Hence proved