If A (3, 2), C (- 3, 2) and D(k, – 1) are the vertices of the quadrilateral ABCD where |AD| = |CD| and the area of Δ ABC is 9 sq. unit. Then the area of quadrilateral ABCD will be:
1. 15 sq. unit
2. 12 sq. unit
3. 36 sq. unit
4. 18 sq. unit
1. 15 sq. unit
2. 12 sq. unit
3. 36 sq. unit
4. 18 sq. unit
Correct Answer – Option 4 : 18 sq. unit
Concept:
1. Let A (x1, y1), B (x2, y2) and C (x3, y3) be the vertices of a Δ ABC, then area of Δ ABC = \(\frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right|\)
2. Distance formula: A(x, y) and B(a, b) be any two points, by distance formula,
\(\rm AB = \sqrt {(x – a)^2 + (y – b)^2}\)
3. (a + b)2 = a2 + 2ab + b2
4. (a – b)2 = a2 – 2ab + b2
Calculation:
Given that, coordinates of point A, C and D are (3, 2), (- 3, 2) and (k, -1) respectively.
Area of Δ ABC = 9 sq. unit
According to the question, |AD| = |CD|
Using distance formula,
\(\sqrt {(3 – k)^2 + (2 + 1)^2}\ =\ \sqrt {(-3 \ -\ k)^2 + (2\ +\ 1)^2}\)
Taking square of both side
⇒ (3 – k)2 + 9 = (3 + k)2 + 9
⇒ 3 + k2 – 6k = 9 + k2 + 6k
⇒ 12k = 0 ⇒ k = 0