Question

hello friends.......
please answer the above problem (2)

only(iii) is enough

Answer 1

Hi there !
Here's the answer:

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• The problem can be solved by using Area of Triangle concept of coordinate geometry.

• If any 3 Points are to collinear, then they will lie on same line
=> they will not form any triangle

• In other words,
Area of ∆ABC = 0

¶ If $A(x_{1} , y_{1})$ , $B(x_{2} , y_{2})$ , $C(x_{3} , y_{3})$ are the points.

=> $\frac{1}{2}[x_{1}(y_{2} - y_{3}) + x_{2}(y_{3} - y_{1}) + x_{3}(y_{1} - y_{2})]$

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SOLUTION :

Given,
$A(x_{1} , y_{1})$ = (K , K),
$B(x_{2} , y_{2})$ = (2 , 3) ,
$C(x_{3} , y_{3})$ = (4 , -1)

Here,
$x_{1} = K$ ; $y_{1} = K$
$x_{2} = 2$ ; $y_{2} = 3$
$x_{3} = 4$ ; $y_{3} = -1$

Substitute values

=> $\frac{1}{2}[K(3 - (-1)) + 2(-1 - K) + 4(K - 3)]= 0$

=> $\frac{1}{2}[3K+ K - 2 - 2K + 4K - 12] = 0$

=> $\frac{1}{2}[6K-14] = 0$

=> $3k - 7 = 0$

=> $k = \frac{7}{3}$

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Hope it helps