Question

$\huge{\underline{\underline{\bf Question}}}$

find a relation between x and y if the points $\sf A(x,y); B(-4,6)$ And $\sf C(-2,3)$ are collinear.

Class - 10th

Chapter - Coordinate geometry

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Answer 1

Given :

• A (x,y)
• B (-4,6)
• C (-2,3)

To Find :

• Relation between x and y points

Formula Applied :

${\Rightarrow \sf{\begin {pmatrix} x_1 & x_2&x_3&x_1\\y_1&y_2&y_3&y_1\end{pmatrix}}}$

${\Rightarrow \sf 0 = \dfrac{1}{2}\Big(x_1y_2+x_2y_3+x_3y_1\Big)-\Big(y_1x_2+y_2x_3+y_3x_1\Big)}$

${\Rightarrow \sf 0 = \dfrac{1}{2}\Big(x_1y_2+x_2y_3+x_3y_1- y_1x_2 - y_2x_3 - y_3x_1\Big)}$

${\Rightarrow \sf 0 \times \dfrac{2}{1}= \Big(x_1y_2-y_3x_1 + x_2y_3 - y_1x_2 + x_3y_1+y_2x_3\Big)}$

${\boxed{\sf 0= \bigg[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\bigg]}}$

Solution :

${\Rightarrow \sf 0= \bigg[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\bigg]}$

${\Rightarrow \sf 0= \bigg[(x)(6-3)+(-4)(3-y)+(-2)(y-6)\bigg]}$

${\Rightarrow \sf 0= \bigg[(6x-3x)+(-12+4y)+(-2y+12)\bigg]}$

${\Rightarrow \sf 0= \bigg[6x-3x{\cancel{-12}}+4y-2y{\cancel{+12}}\bigg]}$

${\Rightarrow \sf 0= \bigg[6x-3x+4y-2y\bigg]}$

${\Rightarrow \sf 0=\bigg[3x+2y\bigg]}$

Now let's determine the relation between x and y co-ordinates,

${\longrightarrow \sf 3x+2y=0}\\\\{\longrightarrow \sf 3x = -2y}\\\\{\longrightarrow \sf x=\dfrac{-2y}{3}}$

Required Answer :

The relation between x and y co-ordinates as by given is $\underline{\sf x = 2y}$

Answer 2

$\huge\underline{\bf{Solution}}$

☆ It is given that points A(x,y); B(-4,6) and C(-2,3) are collinear. We have to find the relation between x and y.

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⠀⠀━━━━━━━━━━━━━━━━━━━━━

We know that, if three points are collinear, it means that all three points are on the same line. To prove that all three points are on same line, we know that area of the given points will be zero.

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Formula of area is given below

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$\bf{\underline{\boxed{\boxed{\small{Area = \dfrac{1}{2} \bigg| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \bigg|}}}}}$

⠀

We have,

• $\sf{x_1 = x, \: y_1 = y}$
• $\sf{x_2 = -4, \: y_2 = 6}$
• $\sf{x_3 = -2, \: y_3 = 3}$

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Putting all the values in formula and we have

⠀⠀⠀⠀❏ Area = 0

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$\tt\dashrightarrow{\dfrac{1}{2} \bigg| x(6 - 3) + (-4)(3 - y) + (-2)(y - 6) \bigg| = 0}$

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$\tt\dashrightarrow{\bigg| x(3) - 12 + 4y - 2y + 12 \bigg| = 0 \times \dfrac{2}{1}}$

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$\tt\dashrightarrow{\bigg| 3x - \cancel{12} + 4y - 2y + \cancel{12} \bigg| = 0}$

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$\tt\dashrightarrow{\bigg| 3x + 2y \bigg| = 0}$

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$\bf\dashrightarrow{\pink{3x + 2y = 0}}$

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$\: \: \: \: \: \: \: \: \: \: \dag{\tiny{\underline{\frak{Hence,\: required\: relation\: is\: 3x + 2y = 0}}}}$