Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Line through the points | Equation of the line | ||
| (A) | \((1, 2)\) and \((3, 6)\) | (I) | \(2x - y = 1\) |
| (B) | \((0, 3)\) and \((3, 0)\) | (II) | \(2x + y = 0\) |
| (C) | \((1, 1)\) and \((4, 7)\) | (III) | \(2x - y = 0\) |
| (D) | \((-1, 2)\) and \((2, -4)\) | (IV) | \(x + y = 3\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
A point \((x, y)\) lies on the line through \((x_1, y_1)\) and \((x_2, y_2)\) exactly when the three points are collinear, that is, when the area determinant is zero:
\( \displaystyle \begin{vmatrix}x & y & 1\\ x_1 & y_1 & 1\\ x_2 & y_2 & 1\end{vmatrix} = 0 \)
(A) \((1, 2)\) and \((3, 6)\)
This matches (III) \(2x - y = 0\).
\(x(2 - 6) - y(1 - 3) + (6 - 6) = 0\)
\(-4x + 2y = 0\), that is \(2x - y = 0\)
(B) \((0, 3)\) and \((3, 0)\)
This matches (IV) \(x + y = 3\).
\(x(3 - 0) - y(0 - 3) + (0 - 9) = 0\)
\(3x + 3y = 9\), that is \(x + y = 3\)
(C) \((1, 1)\) and \((4, 7)\)
This matches (I) \(2x - y = 1\).
\(x(1 - 7) - y(1 - 4) + (7 - 4) = 0\)
\(-6x + 3y + 3 = 0\), that is \(2x - y = 1\)
(D) \((-1, 2)\) and \((2, -4)\)
This matches (II) \(2x + y = 0\).
\(x(2 + 4) - y(-1 - 2) + (4 - 4) = 0\)
\(6x + 3y = 0\), that is \(2x + y = 0\)