Math Olympiad Test: Coordinate Geometry- 2 - Class 9 MCQ

The horizontal axis in the coordinate plane is called the x-axis. The vertical axis is called the y-axis. The point at which the two axes intersect is called the origin.

If P and Q have same abscissa but different ordinates then P and Q have coordinates, as P (a, c ), Q (a, b)
∵ x-coordinate is constant.
∴ P and Q will lie on line parallel to y-axis.

Minimum distance of point (α, b) from x-axis = Perpendicular distance between point and x-axis = |y-coordinate (ordinate) of the point| = |6| = 6

• Coordinates with positive x and negative y lie in the fourth quadrant.
• Point P has x=4 (positive) and y=−3 (negative).
• Thus, P is in quadrant IV.

Solution:

1. Points on the y-axis: Points on the y-axis always have an x-coordinate of 0. 
2. Same absolute value, different signs: If the x-coordinate of M is K, and M and N have the same absolute value but opposite signs, then the x-coordinate of N must be -K. 
3. Distance formula: The distance between two points (x₁, y₁) and (x₂, y₂) is given by √((x₂ - x₁)² + (y₂ - y₁)²). 
4. Applying to this case: Since both points lie on the y-axis, their y-coordinates will be different. Let's assume the y-coordinate of M is 'a' and the y-coordinate of N is 'b'. So, M = (0, a) and N = (0, b). The distance between them is √((0-0)² + (b-a)²) = √(b-a)². However, since the points have the same absolute x-value but opposite signs, one is K and the other is -K. The distance between M(K, 0) and N(-K, 0) is √((-K - K)² + (0 - 0)²) = √(-2K)² = |2K| = 2|K|. 
5. Final Answer: Therefore, the distance between M and N is 2|K|. 

∵ (-3, -2) has (-, -) sign convention.
∴ (-3, -2)/0 Belongs to 3^rd quadrant.

Point A is the reflected point.
Point A will have coordinates, as, abscissa will not change and ordinate will change sign
∴ Reflected point will lie in 2^nd quadrant.

In ΔOAB
OA² = OB² + AB²
= (12)² + (OC)² = (12)² + (5)²
= 169
⇒ OA = √169 = 13 Units

On every point of y-axis, abscissa = 0
∴ x =  0 as the equation of y-axis

Coordinates of point Q ≡ (-3, 5).
Coordinates of point P ≡ (3, 5).
Coordinates of point O ≡ (0, 0).
After plotting ΔPOQ on graph, it can be clearly viewed that ΔPOQ is isosceles.
∴ Area of ΔPOQ = 1/2 × OR × PQ
= 1/2 × 5 × [3- (-3)]
= 1/2 × 5 × 6 = 15 sq. units

Distance between (-24, 10) and (48,10) will be equal to the absolute value of difference between the abscissa of the points as, the points have same ordinate.
∴ Distance = |48 - (-24)| = 72 units 

The difference between the ordinates of points P (3, -6) and Q (-6, 3) can be calculated as follows:

• Identify the ordinates (y-coordinates) of both points:
  • For point P, the ordinate is -6.
  • For point Q, the ordinate is 3.
• Calculate the difference:
  • Difference = Ordinate of Q - Ordinate of P
  • Difference = 3 - (-6) = 3 + 6 = 9.

Therefore, the difference between the ordinates of points P and Q is 9.

Let the coordinates of point of intersection be (x₁, y₁)
∴ x₁ + y₁ = 6
x₁ - y₁ = 2
Solving these two equations, we get
x₁ = 4, and , y₁ = 2
∴ point of intersection  ≡ (4, 2)

In ΔOBC

BC² + OC² = OB²
⇒ (OA)² + (OC)² = OB²
⇒ (10)² + (24)² = OB²
⇒ OB = √676 = 26 units 

After plotting figure, it can be clearly seen that ABC is an isosceles triangle in which,  
AB = (6 - 2) = 4 units
CD = (6 - 0) = 6 units

∴ Ares of ΔABC = 1/2 × AB × CD
= 1/2 × 4 × 6 = 12 sq. units