Question

find the value of a given points are collinear. (2, 3) , (4, a), (6, -3)
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Answer 1

Solution :-

Let the points be :-

A ( 2 , 3 )

B ( 4 , a )

C ( 6 , -3 )

Given that points are collinear. That is, area of triangle ABC 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 :-

$\bullet \sf \ x_1 = 2 \ , \ y_1 = 3 \\\\\bullet \ x_2 = 4 \ , \ y_2 = a \\\\\bullet \ x_3 = 6 \ , \ y_3 =-3$

$\longrightarrow \sf \dfrac{1}{2} \times [ \ 2(a-(-3)) +4(-3-3)+6(3-a)\ ] = 0 \\\\\longrightarrow \dfrac{1}{2} \times [ \ 2(a+3)+4(-6)+18-6a \ ] = 0 \\\\\longrightarrow \dfrac{1}{2} \times [ \ 2a+6-24+18-6a \ ] = 0 \\\\\longrightarrow \dfrac{1}{2} \times [\ -4a-18+18 \ ] = 0 \\\\\longrightarrow \dfrac{1}{2} \times -4a = 0 \\\\\longrightarrow -4a = 0 \\\\\longrightarrow \bf a = 0$

Answer 2

$\huge\underline{\bf \orange{AnSweR :}}$

Let the points be :-

A ( 2 , 3 )

B ( 4 , a )

C ( 6 , -3 )

Given that points are collinear. That is, area of triangle ABC is equal to zero.

Area of triangle=

2/1 ×[ x¹(y² −y³)+x²(y³− y¹)+x³(y¹− y 2) ]

Here :-

∙ x¹=2 ,

y¹=3

∙ x²=4 ,

y²=a

x³=6 ,

y³ = −3

⟶ 2/1 ×[ 2(a−(−3))+4(−3−3)+6(3−a) ]=0

⟶ 2/1×[ 2(a+3)+4(−6)+18−6a ]=0

⟶ 2/1 ×[ 2a+6−24+18−6a ]=0

⟶ 2/1×[ −4a−18+18 ]=0

⟶ 2/1 ×−4a=0

⟶−4a=0

⟶a=0