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Question: The points A\[(k,2-2k)\], B\((-k+1,2k)\) and C\((-4-k,6-2k)\) collinear for (This question has mul...

The points A(k,2−2k)(k,2-2k), B(−k+1,2k)(-k+1,2k) and C(−4−k,6−2k)(-4-k,6-2k) collinear for
(This question has multiple correct options)
A) All values of k
B) k = -1
C) k = 12\dfrac{1}{2}
D) no value of k

Explanation

Solution

Hint: The area of the triangle made by the collinear point should be equal to zero. Solve the equation and we get the value of k.

Complete-step-by-step-Solution:
In this question there are three given points which are collinear. The points are given in terms of k and here we have to find the value of k for which these points are collinear.
As we know that if three points are given as collinear then the area made by them will be equal to zero.
So, we will use this concept here to find the value of k.
Take the coordinates as A(x1,y1),({{x}_{1}},{{y}_{1}}),B(x2,y2)({{x}_{2}},{{y}_{2}})and C(x3,y3)({{x}_{3}},{{y}_{3}}) as A=(k,2−2k)(k,2-2k), B=(−k+1,2k)(-k+1,2k) and C=(−4−k,6−2k)(-4-k,6-2k)
Now, we know that these points are collinear so we will find the area of triangle from these points
So,
Area of triangle through as A(x1,y1),({{x}_{1}},{{y}_{1}}),B(x2,y2)({{x}_{2}},{{y}_{2}})and C(x3,y3)({{x}_{3}},{{y}_{3}}) will be
Area = 12∣[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]∣\dfrac{1}{2}|[{{x}_{1}}({{y}_{2}}-{{y}_{3}})+{{x}_{2}}({{y}_{3}}-{{y}_{1}})+{{x}_{3}}({{y}_{1}}-{{y}_{2}})]|
And for collinear points this area will be equal to zero.
So,
⇒12∣[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]∣=0\Rightarrow \dfrac{1}{2}|[{{x}_{1}}({{y}_{2}}-{{y}_{3}})+{{x}_{2}}({{y}_{3}}-{{y}_{1}})+{{x}_{3}}({{y}_{1}}-{{y}_{2}})]|=0
Now we will put the value of all the coordinate value and we will solve for k
⇒12∣[k((2k)−(6−2k))+(−k+1)((6−2k)−(2−2k))+(−4−k)((2−2k)−(2k))]∣=0 ⇒12∣[k(2k−6+2k)+(−k+1)(6−2k−2+2k)+(−4−k)(2−2k−2k)]∣=0 ⇒12∣[k(4k−6)+(−k+1)(4)+(−4−k)(2−4k)]∣=0 ⇒12∣[(4k2−6k)+(−4k+4)+(−8+16k−2k+4k2)]∣=0 ⇒12∣[(4k2−6k−4k+4−8+16k−2k+4k2)]∣=0 ⇒(8k2+4k−4)=0 ⇒(2k2+k−1)=0 ⇒(2k2+2k−k−1)=0 ⇒(2k(k+1)−1(k+1)=0 ⇒(k+1)(2k−1)=0 ⇒k=−1,k=12 \begin{aligned} & \Rightarrow \dfrac{1}{2}|[k((2k)-(6-2k))+(-k+1)((6-2k)-(2-2k))+(-4-k)((2-2k)-(2k))]|=0 \\\ & \Rightarrow \dfrac{1}{2}|[k(2k-6+2k)+(-k+1)(6-2k-2+2k)+(-4-k)(2-2k-2k)]|=0 \\\ & \Rightarrow \dfrac{1}{2}|[k(4k-6)+(-k+1)(4)+(-4-k)(2-4k)]|=0 \\\ & \Rightarrow \dfrac{1}{2}|[(4{{k}^{2}}-6k)+(-4k+4)+(-8+16k-2k+4{{k}^{2}})]|=0 \\\ & \Rightarrow \dfrac{1}{2}|[(4{{k}^{2}}-6k-4k+4-8+16k-2k+4{{k}^{2}})]|=0 \\\ & \Rightarrow (8{{k}^{2}}+4k-4)=0 \\\ & \Rightarrow (2{{k}^{2}}+k-1)=0 \\\ & \Rightarrow (2{{k}^{2}}+2k-k-1)=0 \\\ & \Rightarrow (2k(k+1)-1(k+1)=0 \\\ & \Rightarrow (k+1)(2k-1)=0 \\\ & \Rightarrow k=-1,k=\dfrac{1}{2} \\\ \end{aligned}
So, on solving the equation we are getting the value of k=−1k=-1 or k=12k=\dfrac{1}{2}

Note: In this type of equation we always need to put the given coordinate in the formula of area of triangle and then we need to solve the acquired polynomial using factorization method for getting value of variables.