Question

find the value of k for which the points (6, -1) (4 ,0) and (2 , k) are collinear ​

Answer 1

Solution :-

Let the points be A ( 6 , -1 ), B ( 4 , 0 ) and C ( 2 , k ).

Given that, the points are collinear.

That is,

Area of ΔABC = 0

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

Here :-

$\bullet \sf \ x_1 = 6 \ , \ y_1 = -1$

$\bullet \sf \ x_2 = 4 \ , \ y_2 = 0$

$\bullet \sf \ x_3 = 2 \ , \ y_3 = k$

$: \implies \sf \dfrac{1}{2} \times [ \ 6(0-k)+4(k-(-1))+2(-1-0) \ ] = 0$

$: \implies \sf \dfrac{1}{2} \times [ \ 6(0-k)+4(k+1)+2(-1-0) \ ] = 0$

$: \implies \sf \dfrac{1}{2} \times [ \ -6k+4k+4-2 \ ] = 0$

$: \implies \sf \dfrac{1}{2} \times [ \ -2k+2 \ ] = 0$

$: \implies \sf -2k+2 = 0$

$: \implies \sf -2k = -2$

$: \implies \bf k=1$

Answer 2

Step-by-step explanation:

Take the points as A,B,C

• A (6,-1)
• B(4,0)
• C(2,k)

Here

$\sf x_1=6,y_1=-1$

$\sf x_2=4,y_2=0$

$\sf x_3=2,y_3=k$

The points are Colinear

${:}\longrightarrow\sf Area \:of\:\triangle ABC=0$

According to Coordinate geometry

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

• Substitute the values

$: \implies \sf \dfrac{1}{2} \times [ \ 6(0-k)+4(k-(-1))+2(-1-0) \ ] = 0$

$: \implies \sf \dfrac{1}{2} \times [ \ 6(0-k)+4(k+1)+2(-1-0) \ ] = 0$

$: \implies \sf \dfrac{1}{2} \times [ \ -6k+4k+4-2 \ ] = 0$

$: \implies \sf \dfrac{1}{2} \times [ \ -2k+2 \ ] = 0$

$: \implies \sf -2k+2 = 0$

$: \implies \sf -2k = -2$

$: \implies \bf k=1$

$\therefore\sf k=1.$