Math Olympiad Test: Co-ordinate Geometry- 2

# Math Olympiad Test: Co-ordinate Geometry- 2

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## 15 Questions MCQ Test International Mathematics Olympiad (IMO) for Class 10 | Math Olympiad Test: Co-ordinate Geometry- 2

Math Olympiad Test: Co-ordinate Geometry- 2 for Class 10 2022 is part of International Mathematics Olympiad (IMO) for Class 10 preparation. The Math Olympiad Test: Co-ordinate Geometry- 2 questions and answers have been prepared according to the Class 10 exam syllabus.The Math Olympiad Test: Co-ordinate Geometry- 2 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- 2 below.
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Math Olympiad Test: Co-ordinate Geometry- 2 - Question 1

### Find the third vertex of a ΔABC if two of its vertices are B (-3, 1), C (0, -2), and its centroid is at the origin.

Detailed Solution for Math Olympiad Test: Co-ordinate Geometry- 2 - Question 1

B  (-3, 1) C (0, -2)
Let the third vertex be A (x, y)
G = (0, 0)
= 0 ⇒ x = 3 and
⇒ y - 1 = 0 ⇒ y = 1
A (3, 1)

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

### The circumcentre of a triangle is (3, 3). If its two vertices are (4, 6) and (0, 4) find the third vertex of the triangle.

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

Let P (3, 3) be the circumcentre of ΔABC, C(x, y) be the third vertex

PA = PB = PC
PA= PB= PC2
PA2 = PC2
⇒ (4 - 3)2 + (6 - 3)2
= (x - 3)2 + (y - 3)2
⇒ 1 + 9 = x2 + 9 - 6x + y2 + 9 - 6y
⇒ x2 + y2 - 6x - 6y + 8 = 0 ...(1)
and PB2 = PC2
⇒ (0 - 3)2 + (4, -3)2 = (x - 3)2 + (y - 3)2
⇒ 9 + 1 = x2 + 9 - 6x + y2 - 6y + 9
Equations (1) and (2) are identical
(x - 3)2 + (y - 3 )2 = 10
So, (6, 2) is the required point.

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

### In which ratio does the point P (1, 2) divides the join of A (-2, 1) and B (7, 4)?

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

Let the ratio be k : 1

⇒ 6k = 3
⇒ k = 3/6 = 1/2

Math Olympiad Test: Co-ordinate Geometry- 2 - Question 4

If (2, -2), (-2, 1) and (5, 2) are vertices of a right angled triangle, then the area of triangle is

Detailed Solution for Math Olympiad Test: Co-ordinate Geometry- 2 - Question 4

Area of triangle
= 1/2 [x1 (y2 - y3) + x2 (y3 -y1) + x3 (y1 - y2
= 1/2 [2 (1 − 2) + (−2) (2 + 2) + 5 (−2 −1)]
= 1/2 [−2−8 −15] = 25/2 = 12.5 sq. units

Math Olympiad Test: Co-ordinate Geometry- 2 - Question 5

The centre of a circle is (4, 5) A(8, 10) is a point on the circumference. Find the other end of diameter of the circle through A.

Detailed Solution for Math Olympiad Test: Co-ordinate Geometry- 2 - Question 5

Let the other end be (x, y)

Math Olympiad Test: Co-ordinate Geometry- 2 - Question 6

The ends of a diameter of a circle have the coordinates (4, 3) and (-4, -3), PQ is another diameter where P has co-ordinate

find the coordinates of Q.

Detailed Solution for Math Olympiad Test: Co-ordinate Geometry- 2 - Question 6

Mid-point of (4, 3) and (-4, -3)

Math Olympiad Test: Co-ordinate Geometry- 2 - Question 7

The vertices of a triangle are A(1, 1), B(-2, -5) and C(2, 2) find the length of the medium through C.

Detailed Solution for Math Olympiad Test: Co-ordinate Geometry- 2 - Question 7

Math Olympiad Test: Co-ordinate Geometry- 2 - Question 8

What is the circumradius of the triangle whose vertices are (2, -2), (8, 6), and (8, -2)?

Detailed Solution for Math Olympiad Test: Co-ordinate Geometry- 2 - Question 8

Let P(x, y) be the circumcentre
PA = PB = PC
⇒ PA2 = PB2
⇒ (x - 2)2 + (y + 2)2
= (x - 8)2 + (y - 6)2
⇒ x2 + 4 - 4x + y2 + 4 - 4y
= x2+ 64 - 16x + y2 + 36 - 12y
⇒ - 4x + 16x + 4y + 12y = 100 - 8
⇒ 12x + 16y = 92 ⇒ 3x+ 4y = 23 ...(1)
and PB2 = PC2
⇒ (x - 8)2 + (y - 6)2 = (x - 8)2 + (y + 2)2
⇒ y2 + 36 - 12y = y2 + 4 + 4y
⇒ 16y = 32 ⇒ y = 2
3x = 23 - 4 ⇒ 2 = 15 ⇒ x = 5
P (5, 2)
∴ PA

Math Olympiad Test: Co-ordinate Geometry- 2 - Question 9

The sum of the square of the distance of a moving point from two fixed points (a, 0) and (-a, 0) is equal to the constant quantity 2c2. Find the equation of its locus.

Detailed Solution for Math Olympiad Test: Co-ordinate Geometry- 2 - Question 9

Here (x -a)2 + (y - 0)2 = 2c2
x2 + a2 - 2ax + y2 = 2c2
and (x + a)2 + (y - 0)2 = 2c2
⇒ x2 + a2 + 2ax + y2 + 2c2
Now x2 + a2 - 2ax + y2 = x2 + a2 + 2ax + y2
⇒ 4ax = 0 ⇒ x = 0 or a = 0
(x - 0)2 + y2 = 2c2
⇒ x0 + y2 = 2c2

Math Olympiad Test: Co-ordinate Geometry- 2 - Question 10

What is the distance of the point (4, 7) from the y-axis?

Detailed Solution for Math Olympiad Test: Co-ordinate Geometry- 2 - Question 10

Distance of the point (4, 7) from (0, y)
= x co-ordinate
= 4

Math Olympiad Test: Co-ordinate Geometry- 2 - Question 11

If the distance between the points (3, 0) and (0, y) is 5 units. y is positive then what is value of y?

Detailed Solution for Math Olympiad Test: Co-ordinate Geometry- 2 - Question 11

(3 - 0)2 + (0 - y)2 = 52
9 + y2 = 25 ⇒ y2 = 16 ⇒ y = ± 4
y is positive, ∴ y = 4

Math Olympiad Test: Co-ordinate Geometry- 2 - Question 12

If the centroid of the triangle is formed by points P(a, b), Q(b, c) and R(c, a) is at the origin. What is the value of a + b +c?

Detailed Solution for Math Olympiad Test: Co-ordinate Geometry- 2 - Question 12

Math Olympiad Test: Co-ordinate Geometry- 2 - Question 13

If the point P(-1, 2) divides externally the line segment joining A(2, 5) and B in the ratio 3 : 4. What is the co-ordinate of point B?

Detailed Solution for Math Olympiad Test: Co-ordinate Geometry- 2 - Question 13

Math Olympiad Test: Co-ordinate Geometry- 2 - Question 14

The points A(3, 1), B(0, 4), C(-3, 1), D(0, -2) are vertices of a

Detailed Solution for Math Olympiad Test: Co-ordinate Geometry- 2 - Question 14

The given points are A (0,-2), B(3,1), C(0,4) and D(-3,1)

Thus, diagonal PR = diagonal QS
Therefore, the given points from a square.

Math Olympiad Test: Co-ordinate Geometry- 2 - Question 15

If the point (x, 4) lies on a circle whose center is at the origin and radius is 5, What is the value of x?

Detailed Solution for Math Olympiad Test: Co-ordinate Geometry- 2 - Question 15

OP = 5
⇒ OP2 = 25
⇒ (x - 0)2 + (4 - 0)2 = 25
⇒ x2 + 16 = 25
⇒ x2 = 9 ⇒ x = ± 3

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