Question

Show that the point (8 ,1) ,(3 ,-4) , (2,-5) are collineas

Answer 1

Solution :-

Let the points be A ( 8 , 1 ), B ( 3 , 4 ) and C ( 2 , -5 ).

We need to show that these points are collinear.

That is,

Area of ΔABC is equal to zero .

$\boxed{\bf Area \ of \ triangle = \dfrac{1}{2} \times [ \ x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}$

Here :-

$\bullet \sf \ x_1=8 \ , \ y_1= 1 \\\\\bullet \ x_2 = 3 \ , \ y_2 = -4 \\\\\bullet \ x_3 = 2 \ , \ y_3 = -5$

$\longrightarrow \sf \dfrac{1}{2} \times [ \ 8(-4-(-5)) +3 (-5-1) +2(1-(-4)) \ ] \\\\\longrightarrow \dfrac{1}{2} \times [ \ 8(-4+5) +3(-5-1)+2(1+4) \ ] \\\\\longrightarrow \dfrac{1}{2} \times [ \ (8 \times 1 ) +( 3\times -6 ) + ( 2 \times 5 ) \ ] \\\\\longrightarrow \dfrac{1}{2} \times [ \ 8-18+10 \ ] \\\\\longrightarrow \dfrac{1}{2} \times [ \ -10+10 \ ] \\\\\longrightarrow \dfrac{1}{2} \times 0 \\\\\longrightarrow 0$

∴ The given points are collinear.

Answer 2

Answer:

✯ Given :-

• (8 , 1), (3 , - 4), (2 , - 5)

✯ To Show:-

• Point (8 , 1), (3 , - 4), (2 , - 5) are collinear.

✯ Solution :-

» Let,A(x₁ = 8, y₁ =1),B(x₂ = 3, y₂ = - 4) and C(x₃ = 2, y₃ = -5)

be the points.

➔ Now, we know that,

$\boxed{\bold{\large{✮\: \sf{x_1} (y₂ - y₃)\: +\: x₂ (y₃ - \sf{y_1})\: +\: x₃ (\sf{y_1} - y₂)\: ✮}}}$

According to the question by using the formula we get,

⇒ 8(- 4 + 5) + 3(- 5 - 1) + 2(1 + 4)

⇒ - 32 + 40 - 15 - 3 + 2 + 8

➠ 0

$\therefore$ The given points are collinear.

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