Question

Find the value of K for which the points are collinear
(K,K),(2,3)and (4,-1)

Answer 1

Answer:-

K = 14/6

Given :-

The points given

( k, k) , ( 2,3 ) and ( 4 , -1)

To show :-

That these points are collinear.

Solution :-

Collinear points :- The points which lies on the same line is called collinear points.

Condition for collinearity :- Area of triangle = 0

Area of triangle is given by :-

★ $\boxed{\sf{\dfrac{1}{2} x_1 (y-2 -y_3 ) +x_2 (y_3 - y_1 ) + x_3 ( y_1 - y_2 )= 0}}$

Let,

$x_1 = k,x_2 = 2 , x_3 = 4 \\ y_1 = k, y_2 = 3 , y_3 = -1$

Now, put the given values,

→$\dfrac{1}{2} k[3 - (-1) ] +2( -1 - k) + 4 ( k - 3) = 0$

→$\dfrac{1}{2} k (4 ) + 2 (-1-k) + 4 (k-3) = 0$

→$\dfrac {1}{2} 4k -2 -2k +4k -12 = 0$

→$\dfrac{1}{2} 8k -2k -2 -12 = 0$

→$\dfrac{1}{2} 6k -14=0$

→$6k -14 = 0$

→$6k =14$

→$k = \dfrac{14}{6}$

Therefore if k = 14/6 then the given points are collinear.

Answer 2

Answer:

Answer

Step-by-step explanation:

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