Question

If the three points (-3,7)(a,1)(-3,2) are collinear then the value of a is​

Answer 1

SOLUTION :-

Given that,

The points ( -3 , 7 ), ( a , 1 ) and ( -3 , 2 ) are collinear.

Let the points be,

A = ( -3 , 7 )

B = ( a , 1 )

C = ( -3 , 2 )

Given that, the points are collinear.

That is,

Area of triangle ABC = 0

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

Here,

$\bullet\sf \ x_1= -3 \ , \ y_1= 7 \\\\\bullet \ x_2= a \ , \ y_2= 1\\\\\bullet\sf \ x_3= -3 \ , \ y_3= 2$

$\longrightarrow\sf \dfrac{1}{2} \times [ \ -3(1-2)+a(2-7)+-3(7-1) \ ]=0 \\\\\longrightarrow \sf \dfrac{1}{2} \times [ \ (-3 \times -1 )+(a \times -5 )+(-3\times 6 ) \ ] = 0 \\\\\longrightarrow \dfrac{1}{2} \times [ \ 3-5a-18 \ ] = 0 \\\\\longrightarrow \dfrac{1}{2}\times [ \ -15-5a \ ] = 0 \\\\\longrightarrow -15-5a = 0 \\\\\longrightarrow -5a= 15 \\\\\longrightarrow\bf a =- 3$

Value of a = -3

Answer 2

SOLUTION :-

Given that,

The points ( -3 , 7 ), ( a , 1 ) and ( -3 , 2 ) are collinear.

Let the points be,

A = ( -3 , 7 )

B = ( a , 1 )

C = ( -3 , 2 )

Given that, the points are collinear.

That is,

Area of triangle ABC = 0

\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) \ ]}

Area of triangle=

21 ×[ x 1(y2−y 3 )+x 2

(y 3 −y 1 )+x 3 (y −y 2 ) ]

Here,

\begin{lgathered}\bullet\sf \ x_1= -3 \ , \ y_1= 7 \\\\\bullet \ x_2= a \ , \ y_2= 1\\\\\bullet\sf \ x_3= -3 \ , \ y_3= 2\end{lgathered} x 1

=−3 , y 1

=7 x 2

=a , y 2

=1 x 3

=−3 , y 3=2

\begin{lgathered}\longrightarrow\sf \dfrac{1}{2} \times [ \ -3(1-2)+a(2-7)+-3(7-1) \ ]=0 \\\\\longrightarrow \sf \dfrac{1}{2} \times [ \ (-3 \times -1 )+(a \times -5 )+(-3\times 6 ) \ ] = 0 \\\\\longrightarrow \dfrac{1}{2} \times [ \ 3-5a-18 \ ] = 0 \\\\\longrightarrow \dfrac{1}{2}\times [ \ -15-5a \ ] = 0 \\\\\longrightarrow -15-5a = 0 \\\\\longrightarrow -5a= 15 \\\\\longrightarrow\bf a =- 3\end{lgathered}

⟶ 21

×[ −3(1−2)+a(2−7)+−3(7−1) ]=0

⟶ 21×[ (−3×−1)+(a×−5)+(−3×6) ]=0

⟶ 21 ×[ 3−5a−18 ]=0

⟶ 21×[ −15−5a ]=0

⟶−15−5a=0

⟶−5a=15

⟶a=−3

Value of a = -3