Math Olympiad Test: Co-ordinate Geometry- 3

# Math Olympiad Test: Co-ordinate Geometry- 3

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## 10 Questions MCQ Test International Mathematics Olympiad (IMO) for Class 10 | Math Olympiad Test: Co-ordinate Geometry- 3

Math Olympiad Test: Co-ordinate Geometry- 3 for Class 10 2022 is part of International Mathematics Olympiad (IMO) for Class 10 preparation. The Math Olympiad Test: Co-ordinate Geometry- 3 questions and answers have been prepared according to the Class 10 exam syllabus.The Math Olympiad Test: Co-ordinate Geometry- 3 MCQs are made for Class 10 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Math Olympiad Test: Co-ordinate Geometry- 3 below.
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Math Olympiad Test: Co-ordinate Geometry- 3 - Question 1

### Find the area (i n square units) of the triangle whose vertices are (a, b + c), (a, b – c) and (–a, c).

Detailed Solution for Math Olympiad Test: Co-ordinate Geometry- 3 - Question 1

Area of triangle

Math Olympiad Test: Co-ordinate Geometry- 3 - Question 2

### Points (6, 8), (3, 7), (–2, –2) and (1, –1) are joined to form a quadrilateral. What will be the structure of quadrilateral?

Detailed Solution for Math Olympiad Test: Co-ordinate Geometry- 3 - Question 2

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.

Math Olympiad Test: Co-ordinate Geometry- 3 - Question 3

### Four vertices of a parallelogram taken in order are (–3, –1), (a, b), (3, 3) and (4, 3). What will be the ratio of a and b?

Detailed Solution for Math Olympiad Test: Co-ordinate Geometry- 3 - Question 3

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,

Math Olympiad Test: Co-ordinate Geometry- 3 - Question 4

Find the area of the quadrilateral, the coordinates of whose angular points taken in order are (1, 1), (3, 4), (5, –2) and (4, –7).

Detailed Solution for Math Olympiad Test: Co-ordinate Geometry- 3 - Question 4

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

Math Olympiad Test: Co-ordinate Geometry- 3 - Question 5

If the points (a, 0), (0, b) and (1, 1) are collinear then which of the following is true?

Detailed Solution for Math Olympiad Test: Co-ordinate Geometry- 3 - Question 5

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,

Math Olympiad Test: Co-ordinate Geometry- 3 - Question 6

The points (1, 1), (–1, 5), (7, 9) and (9, 5) taken in such order that it will form a

Detailed Solution for Math Olympiad Test: Co-ordinate Geometry- 3 - Question 6

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.

Math Olympiad Test: Co-ordinate Geometry- 3 - Question 7

The coordinates of the mid-points of the sides of a triangle are (4, 2), (3, 3) and (2, 2). What will be the coordinates of the centroid of the triangle?

Detailed Solution for Math Olympiad Test: Co-ordinate Geometry- 3 - Question 7

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 =

Math Olympiad Test: Co-ordinate Geometry- 3 - Question 8

Three points A(1, –2), B(3, 4) and C(4, 7) form

Detailed Solution for Math Olympiad Test: Co-ordinate Geometry- 3 - Question 8

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.

Math Olympiad Test: Co-ordinate Geometry- 3 - Question 9

Find the area of triangle whose vertices are (t, t – 2), (t + 2, t + 2) and (t + 3, t).

Detailed Solution for Math Olympiad Test: Co-ordinate Geometry- 3 - Question 9

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

Math Olympiad Test: Co-ordinate Geometry- 3 - Question 10

Area of quadrilateral formed by the vertices (–1, 6), (–3, –9), (5, –8) and (3, 9) is _______ (sq. units).

Detailed Solution for Math Olympiad Test: Co-ordinate Geometry- 3 - Question 10

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.

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