find a relation between x and y if the points And are collinear. Cla

1.	find a relation between x and y if the points And are collinear. Class - 10th Chapter - Coordinate geometry _________________________NO COPY-PASTE ❌NO SPAM ❌JUNIORS DON'T ANSWER ⚠️__________________________Mark my words I'm already in a bad mood ⚠️⚠️⚠️
Answer» ONG>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 $\underline{\sf x = 2y}$

find a relation between x and y if the points And are collinear. Class - 10th Chapter - Coordinate geometry _NO COPY-PASTE ❌NO SPAM ❌JUNIORS DON'T ANSWER ⚠️__Mark my words I'm already in a bad mood ⚠️⚠️⚠️

Answer» ONG>GIVEN :

To Find :

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 $\underline{\sf x = 2y}$