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Question: Let \( A\left( {{x}_{1}},{{y}_{1}} \right),B\left( {{x}_{2}},{{y}_{2}} \right) \) and \( C\left( {{x...

Let A(x1,y1),B(x2,y2)A\left( {{x}_{1}},{{y}_{1}} \right),B\left( {{x}_{2}},{{y}_{2}} \right) and C(x3,y3)C\left( {{x}_{3}},{{y}_{3}} \right) be three points such that abscissas and ordinates from 2 different A.P.’s. Then these points;
A. Form an equilateral triangle
B. Are collinear
C. Are concyclic
D. None of these

Explanation

Solution

To find that what these point will form we will simply write the coordinated in the form of A.P. and then find out the determinant of these three points, if it come out zero then these three points are scalar, otherwise they can form triangle

Complete step by step answer:
Moving ahead with the question in step wise manner,
As according to question x1,x2,x3{{x}_{1}},{{x}_{2}},{{x}_{3}} and y1,y2,y3{{y}_{1}},{{y}_{2}},{{y}_{3}} are in A.P. let the common difference of x1,x2,x3{{x}_{1}},{{x}_{2}},{{x}_{3}} is d1{{d}_{1}} and that of y1,y2,y3{{y}_{1}},{{y}_{2}},{{y}_{3}} be d2{{d}_{2}} . So as x1,x2,x3{{x}_{1}},{{x}_{2}},{{x}_{3}} and y1,y2,y3{{y}_{1}},{{y}_{2}},{{y}_{3}} are in A.P. so we can say that;
x1=x1, x2=x1+d1, x3=x1+2d1 \begin{aligned} & {{x}_{1}}={{x}_{1}}, \\\ & {{x}_{2}}={{x}_{1}}+{{d}_{1}}, \\\ & {{x}_{3}}={{x}_{1}}+2{{d}_{1}} \\\ \end{aligned}
And,
y1=y1, y2=y1+d2, y3=y1+2d2 \begin{aligned} & {{y}_{1}}={{y}_{1}}, \\\ & {{y}_{2}}={{y}_{1}}+{{d}_{2}}, \\\ & {{y}_{3}}={{y}_{1}}+2{{d}_{2}} \\\ \end{aligned}
Now to find out what three points will form let us find out the determinant of these three points. As we know that if these three points of triangle, or concyclic, then the determinant of these three points will be non-zero, and if it comes out to be zero then the three points are collinear. So let us first write these three points in determinant form; so we will get;[1x1y1 1x1+d1y1+d2 1x1+2d1y1+2d2 ]\left[ \begin{matrix} 1 & {{x}_{1}} & {{y}_{1}} \\\ 1 & {{x}_{1}}+{{d}_{1}} & {{y}_{1}}+{{d}_{2}} \\\ 1 & {{x}_{1}}+2{{d}_{1}} & {{y}_{1}}+2{{d}_{2}} \\\ \end{matrix} \right]
Now let us simplify using the properties of determinant, let us function in Row-3 subtract it with row-2 i.e. R3→R3−R2R_3\to R_3-R_2
So we will get the determinant;

& \left[ \begin{matrix} 1 & {{x}_{1}} & {{y}_{1}} \\\ 1 & {{x}_{1}}+{{d}_{1}} & {{y}_{1}}+{{d}_{2}} \\\ 1-1 & {{x}_{1}}+2{{d}_{1}}-\left( {{x}_{1}}+{{d}_{1}} \right) & {{y}_{1}}+2{{d}_{2}}-\left( {{y}_{1}}+{{d}_{2}} \right) \\\ \end{matrix} \right] \\\ & \left[ \begin{matrix} 1 & {{x}_{1}} & {{y}_{1}} \\\ 1 & {{x}_{1}}+{{d}_{1}} & {{y}_{1}}+{{d}_{2}} \\\ 0 & {{d}_{1}} & {{d}_{2}} \\\ \end{matrix} \right] \\\ \end{aligned}$$ Now let us add row 3 in row 1 i.e. $ R_1\to R_1+R_3 $ , so that we will get one complete row 1 same as row 2, so we will get; $$\left[ \begin{matrix} 1 & {{x}_{1}}+{{d}_{1}} & {{y}_{1}}+{{d}_{2}} \\\ 1 & {{x}_{1}}+{{d}_{1}} & {{y}_{1}}+{{d}_{2}} \\\ 0 & {{d}_{1}} & {{d}_{2}} \\\ \end{matrix} \right]$$ Now as we can clearly see that row 1 is the same as row 2, so by the property of determinant, which says that whenever two rows or columns are the same, then the value of that determinant is zero. So it will give us zero. Which means we got the determinant of three coordinates zero. Which means that these coordinates are collinear. **So, the correct answer is “Option B”.** **Note:** In determinants if two rows or columns are the same then the value of that determinant is zero. If the value of the determinant will not come to zero, then we will go to check the angle between two adjacent sides, if they come out to be $ {{60}^{\circ }} $ then it is an equilateral triangle otherwise not.