Consider the following statements:
(A) The points \((1, 1)\), \((3, 5)\) and \((-1, -3)\) are collinear.
(B) The area of the triangle with vertices \((0, 0)\), \((3, 5)\) and \((4, 1)\) is \(\displaystyle\frac{17}{2}\) square units.
(C) The area of the triangle with vertices \((0, 0)\), \((2, 0)\) and \((0, 6)\) is 6 square units.
(D) The points \((2, 3)\), \((4, k)\) and \((6, 7)\) are collinear only when \(k = 4\).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A), (B) and (C) only
The area of the triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\) is the absolute value of
\( \displaystyle \frac{1}{2}\big[x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\big] \)
which is the expansion of \(\frac{1}{2}\begin{vmatrix}x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1\end{vmatrix}\). Three points are collinear exactly when this is zero.
For (A):
\(1(5 + 3) + 3(-3 - 1) + (-1)(1 - 5) = 8 - 12 + 4 = 0\)
For (B):
\( \displaystyle \frac{1}{2}\big[0 + 3(1 - 0) + 4(0 - 5)\big] = \frac{1}{2}(3 - 20) = -\frac{17}{2} \)
For (C):
\( \displaystyle \frac{1}{2}\big[0 + 2(6 - 0) + 0\big] = 6 \)
For (D):
\(2(k - 7) + 4(7 - 3) + 6(3 - k) = 0\)
\(-4k + 20 = 0\)
\(k = 5\)
So the points in (A) give area 0, the areas in (B) and (C) are \(\frac{17}{2}\) and 6, and the points in (D) are collinear only for \(k = 5\).
Checking the statements
(A) \((1, 1)\), \((3, 5)\), \((-1, -3)\) collinear — Correct. The area expression is \(8 - 12 + 4 = 0\).
(B) Area \(\frac{17}{2}\) square units — Correct. \(\frac{1}{2}|3 - 20| = \frac{17}{2}\).
(C) Area 6 square units — Correct. \(\frac{1}{2}|2 \times 6| = 6\).
(D) Collinear only when \(k = 4\) — Incorrect. \(-4k + 20 = 0\) gives \(k = 5\).