Question

If three points A,B,C are collinear, then which of the following is true

a.AB + BC > AC
b.AB + BC = AC
c.AB + BC ≠ AC
d.AB + BC < AC
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Answer 1

Question: If three points A, B and C are Collinear, then which of the following is true?

• AB + BC > AC
• AB + BC = AC
• AB + BC ≠ AC
• AB + BC < AC

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AnswEr : If three given Points A, B and C are Collinear, then 'AB + BC = AC'. Option ( b ) is correct.

• Collinear Points are the points which lie on a same line.
• Non – Collinear Points are the points which do not lie on a same line.

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✇ If we've to find the Collinear Points then there are three Formulas to find out the Collinear Points which are Given by —

⠀⠀( I ) Distance Formula

⠀⠀( II ) Area of Triangle

⠀⠀( III ) Slope Formula

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❍ Basically, Distance Formula is used to find the distance b/w any two given Points:

• $\sf{\sqrt{\Big\{x_2 - x_1\Big\}^2 + \Big\{y_2 - y_1\Big\}^2}}$

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❍ Area of Triangle, when the area of Triangle formed by three point is Zero. Formula:

• $\sf{\dfrac{1}{2} \Bigg[x_1(y_2 - y_3) +x_2(y_3 - y_1) + x_3(y_1 - y_2)\Bigg]}$

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❍ Slope formula measures the line segment step by step, formula:

• $\sf{Q = \Bigg(\dfrac{y_2 - y_1}{x_2 - x_1}\Bigg)}$

Answer 2

$\red{\bigstar}$ Q U E S T I O N

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• If three points A, B and C are collinear, then which of the following is true?

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$\blue{\bigstar}$ G I V E NㅤO P T I O N S

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• (a) AB + BC > AC

• (b) AB + BC = AC

• (c) AB + BC ≠ AC

• (d) AB + BC < AC

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$\purple{\bigstar}$ A N S W E R

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• Option (b) AB + BC = AC is correct!

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$\green{\bigstar}$ M O R EㅤT OㅤK N O W

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• Collinear points are points lying on same line.

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• The distance between $\bf A(x_{1},\:y_{1})$ and $\bf B(x_{2},\:y_{2})$ is :: $\bf \sqrt{\Big(x_{2} - x_{1}\Big)^2 + \Big(y_{2} - y_{1}\Big)^2}$

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• Distance of a point P(x, y) from the origin is :: $\bf \sqrt{x^2 + y^2}$

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• The coordinates of the point P(x, y) which divides the line segment joining the points $\bf A(x_{1},\:y_{1})$ and $\bf B(x_{2},\:y_{2})$ internally in the ratio $\bf m_{1}\::\:m_{2}$ are :: $\bf \bigg\lgroup \dfrac{m_{1}x_{2} + m_{2}x_{1}}{m_{1} + m_{2}},\:\dfrac{m_{1}y_{2} + m_{2}y_{1}}{m_{1} + m_{2}} \bigg\rgroup$

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• The mid - point of the line segment joining the points $\bf A(x_{1},\:y_{1})$ and $\bf B(x_{2},\:y_{2})$ is :: $\bf \bigg\lgroup \dfrac{x_{1} + x_{2}}{2},\:\dfrac{y_{1} + y_{2}}{2}\bigg\rgroup$

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• The area of triangle formed by points $\bf (x_{1},\:y_{1})$, $\bf (x_{2},\:y_{2})$ and $\bf (x_{3},\:y_{3})$ is the numerical value of the expression :: $\small\bf\dfrac{1}{2}\Big[x_{1}\big(y_{2} - y_{3}\big) + x_{2}\big(y_{3} - y_{1}\big) + x_{3}\big(y_{1} - y_{2}\big)\Big]$

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