Math Olympiad Test: Coordinate Geometry- 3 - Free MCQ with solutions

Area of triangle

Let the points be A(6, 8), B(3, 7), C(–2, –2) and D(1, –1)

Now, AB = 

Also, AC = 
BD = 
Since, AB = DC and BC = DA and AC ≠ BD
∴ It is a parallelogram.

Let points be A (–3, –1), B(a, b), C(3, 3) and D(4, 3). So, coordinates of the mid-point of AC = coordinates of the mid-point of BD [∵ In parallelogram, diagonals bisect each other]

⇒ a = – 4 and b = –1
Now, 

Let ABCD be a quadrilateral with vertices
A(1, 1), B(3, 4), C(5, –2) and D(4, –7)

Area of quadrilateral ABCD
= Area of ΔABC + Area of ΔADC...(i)
Now, area of ΔABC

Similarly, area of ΔADC =
= (1/2) × 23 = 11.5 sq.units
Hence, area of quadrilateral ABCD
= 9 + 11.5 = 20.5 sq. units

Since, the given points A(a, 0), B(0, b) and C(1, 1) are collinear
∴ Area of ΔABC = 0
⇒ 1/2|a(b − 1) + 0(1− 0) + 1(0 − b)| = 0
⇒ ab – a – b = 0 ⇒ a + b = ab
Dividing both sides by ab, we get,

Let the points be A(1, 1), B(–1, 5), C(7, 9) and D(9, 5)

Now, by using distance formula, we get
AB = CD, BC = DA and AC = BD
∴ ABCD forms a rectangle.

Let PQR be a triangle and A (4, 2), B (3, 3) and C (2, 2) be the mid-points of sides PQ, PR and QR respectively.
Now, G is the centroid of triangle PQR. Also, G (x, y) is the centroid of triangle formed by joining A, B and C.


⇒ G = 

Distance from A to B = 
Distance from B to C = 
Distance from C to A = 
Since, AB + BC = AC
∴ Points lie on a straight line.

Let A (t, t – 2), B(t + 2, t + 2) and C(t + 3, t) be the vertices of the given triangle. Then,
Area of ΔABC =1/2|{t(t + 2 − t ) + (t + 2)(t − t + 2) + (t + 3)(t – 2 – t – 2)}|
⇒ Area of ΔABC =1/2|{2t + 2t + 4 − 4t − 12} |=| −4 |= 4sq. units

Let A(–1, 6), B(–3, –9), C(5, –8) and D(3, 9) are the vertices of quadrilateral ABCD.

Then, Area of quadrilateral ABCD
= Area of ΔABC + Area of ΔACD ...(i)
Area of (ABCD) = 59 + 37 = 96 sq. units.