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Question: If (3k-1, k-2), (k, k -4) and (k-1, k-2) are collinear points. What is the value of K?...

If (3k-1, k-2), (k, k -4) and (k-1, k-2) are collinear points. What is the value of K?

Explanation

Solution

Hint:Assume the given points as: A(3k-1, k-2), B(k, k -4) and C(k-1, k-2). Find the slope of line segment AB and BC by using the formula: slope=ΔyΔx=y1−y2x1−x2slope=\dfrac{\Delta y}{\Delta x}=\dfrac{{{y}_{1}}-{{y}_{2}}}{{{x}_{1}}-{{x}_{2}}}. Then equate the two slopes equal to each other and simplify the expression to get the answer.

Complete step-by-step answer:
In geometry, collinear points are the set of points which lie on a single straight line.
Now, let us come to the question. We have been given that, (3k-1, k-2), (k, k -4) and (k-1, k-2) are collinear points. Let us assume these points as A, B and C respectively. Since, all the three points lie on the same line then their slopes must be equal. Therefore,
We know that, slope of line segment ΔyΔx=y1−y2x1−x2\dfrac{\Delta y}{\Delta x}=\dfrac{{{y}_{1}}-{{y}_{2}}}{{{x}_{1}}-{{x}_{2}}}
For the line segment AB, assuming:
x1=k, x2=3k−1 y1=k−4, y2=k−2 \begin{aligned} & {{x}_{1}}=k,\text{ }{{x}_{2}}=3k-1 \\\ & {{y}_{1}}=k-4,\text{ }{{y}_{2}}=k-2 \\\ \end{aligned}
∴\therefore Slope of line segment AB =k−4−(k−2)k−(3k−1)=−21−2k=\dfrac{k-4-(k-2)}{k-(3k-1)}=\dfrac{-2}{1-2k}
Now, for the line segment BC, assuming:
x1=k, x2=k−1 y1=k−4, y2=k−2 \begin{aligned} & {{x}_{1}}=k,\text{ }{{x}_{2}}=k-1 \\\ & {{y}_{1}}=k-4,\text{ }{{y}_{2}}=k-2 \\\ \end{aligned}
∴\therefore Slope of line segment BC =K−4−(k−2)k−(k−1)=−21=\dfrac{K-4-(k-2)}{k-(k-1)}=\dfrac{-2}{1}
On equating the slopes of AB and BC equal to each other, we get,
−21−2k=−21\dfrac{-2}{1-2k}=\dfrac{-2}{1}
By cross-multiplication, we get,
−2=−2+4k ⇒4k=−2+2 ⇒4k=0 ⇒k=0 \begin{aligned} & -2=-2+4k \\\ & \Rightarrow 4k=-2+2 \\\ & \Rightarrow 4k=0 \\\ & \Rightarrow k=0 \\\ \end{aligned}
Now, let us check, what happens when we substitute (k = 0) in the above points A, B and C?
The three points become A(-1, -2), B(0, -4) and C(-1, -2). Here, we can see that points A and C become the same. Therefore, we have only two points, A and B. We know that the two points are always collinear. Therefore, the conclusion is that the given points are always collinear for (k = 0).

Note: Here, in this question we have equated the slope of line segment AB and BC. Now, if k is not equal to zero then the three points will never be collinear because the slope of line AC is zero and none of the two other two lines have their slopes zero. One important thing is that, we cannot use distance formulas because we don’t know properly which point is in the middle.