RS Aggarwal Test: Coordinate Geometry - Grade 10 MCQ

The mid-point of a line segment can be found using the formula:

• The x-coordinate of the mid-point is the average of the x-coordinates of the endpoints.
• The y-coordinate of the mid-point is the average of the y-coordinates of the endpoints.

For points A (-2, 8) and B (-6, -4):

• Mid-point x-coordinate: ((-2 + -6) / 2 = -4)
• Mid-point y-coordinate: ((8 + -4) / 2 = 2)

Thus, the mid-point is (-4, 2).

∵ AOB is a right triangle.
∴ Mid-point of AB is equidistant from A, O and B. 

To find the distance of point P(2, 3) from the x-axis:

- The distance from a point (x, y) to the x-axis is the absolute value of the y-coordinate.
- For point P(2, 3), the y-coordinate is 3.
- Therefore, the distance of point P from the x-axis is 3.
- Hence, the correct answer is B: 3.

So, a/3 = -4 ⇒ a = -12

∵ Points are collinear.
∴ (k + 1) (3 - 2k) + (2k, + 1) (2k - 1) + (2k + 2) (1 - 3) = 0
⇒ 3k+3 - 2k² - 2k + 4k² - 1 - 4k - 4 = 0 ⇒ 2k² - 3k - 2 = 0
⇒ 2k² - 4k + k - 2 = 0
⇒ 2k(k - 2) + 1(k - 2) = 0
⇒ (2k + 1) (k - 2) = 0

To calculate the distance of the point A(-5, 6) from the origin (0, 0), we use the distance formula:

Distance = √((x₂ - x₁)² + (y₂ - y₁)²)

Substituting the values:

• x₁ = 0, y₁ = 0
• x₂ = -5, y₂ = 6

We get:

Distance = √((-5 - 0)² + (6 - 0)²)

Simplifying further:

Distance = √((-5)² + 6²)
Distance = √(25 + 36)
Distance = √61

Thus, the correct answer is:

(a) √61 units