Question

Show that the point A[4.2) B(7,5) and C(9,7) are three points lie on a same line​

Answer 1

Correct Question:

• Show that the points A(4, 2), B(7, 5) and C(9, 7) are three point lie on a same line.

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❍ Let the given points be A(4, 2), B(7, 5) and C(9, 7).

◗ To find out distance b/w two given points the formula is given by,

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$\bf{\dag}\:\boxed{\sf {D_{\:(distance)}= \sqrt{\Big(x_2 - x_1 \Big)^2 + \Big(y_2 - y_1 \Big)^2}}}$

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$\underline{\bf{\dag} \:\mathfrak{Substituting \: Values \: in \: the \: formula \: :}}$

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TAKING AB :

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$:\implies\sf AB = \sqrt{\Big(7 - 4 \Big)^2 + \Big(5 - 2 \Big)^2} \\\\\\:\implies\sf AB = \sqrt{3^2 + 3^2} \\\\\\:\implies\sf AB = \sqrt{9 + 9} \\\\\\:\implies\sf AB = \sqrt{18} \\\\\\:\implies{\underline{\boxed{\frak{\pink{AB = 3\sqrt{2} \: units}}}}}\:\bigstar$

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TAKING BC :

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$:\implies\sf BC = \sqrt{\Big(9 - 7 \Big)^2 + \Big(7 - 5 \Big)^2} \\\\\\:\implies\sf BC = \sqrt{2^2 + 2^2} \\\\\\:\implies\sf BC = \sqrt{4 + 4} \\\\\\:\implies\sf BC = \sqrt{8} \\\\\\:\implies{\underline{\boxed{\frak{\pink{BC = 2\sqrt{2} \: units}}}}}\:\bigstar$

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TAKING AC:

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$:\implies\sf AC = \sqrt{\Big(9 - 4 \Big)^2 + \Big(7 - 2 \Big)^2} \\\\\\:\implies\sf AC = \sqrt{5^2 + 5^2} \\\\\\:\implies\sf AC = \sqrt{25 + 25} \\\\\\:\implies\sf AC = \sqrt{50} \\\\\\:\implies{\underline{\boxed{\frak{\pink{AC = 5\sqrt{2} \: units}}}}}\:\bigstar$

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◗ Here, we can see that,

$:\implies\sf 3\sqrt{2} + 2\sqrt{2} = 5\sqrt{2} \\\\\\:\implies{\underline{\boxed{\sf{\purple{AB + BC + AC}}}}}\:\bigstar$

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• NOTE: Collinear points lie on the straight line.

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$\therefore\:{\underline{\sf{Hence, \ given \ points \ are \ \bf{Collinear}.}}}$⠀⠀

Answer 2

Answer:

Question :-

Show that the points A(4, 2), B(7, 5) and C(9, 7) are three point lie on a same line.

Required Answer :-

Use Distance formula for AB,BC,AC

$\sf \: D_{distance} = \big \sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2}$

AB

$\sf \: D \:= \sqrt{(7 - 4) {}^{2}+{5 - 2} {}^{2} }$

$\sf \: D = \sqrt{ {3}^{2} + {3}^{2} }$

$\sf \: D = \sqrt{9 + 9}$

$\sf \: D = \sqrt{18}$

$\sf \: D = 3 \sqrt{2} \: units$

AC

$\sf \: D \:= \sqrt{(9 - 7) {}^{2}+{7-5} {}^{2} }$

$\sf \: D = \sqrt{ {2}^{2} + {2}^{2} }$

$\sf \: D = \sqrt{4 + 4}$

$\sf \: D = \sqrt{8}$

$\sf \: D = 2 \sqrt{2} \: units$

AC

$\sf \: D \:= \sqrt{(9 - 4) {}^{2}+{7 - 2} {}^{2} }$

$\sf \: D = \sqrt{ {5}^{2} + {5}^{2} }$

$\sf \: D = \sqrt{25 + 25}$

$\sf \: D = \sqrt{50}$

$\sf \: D = 5 \sqrt{2} \: units$

Now,

AB + BC = AC

2√2 + 3√2 = 5√2