Find the ratio in which the line segment joining (2, -3) and (5, 6) is divided by x-axis

To find the ratio in which the x-axis divides the line segment joining the points (2, -3) and (5, 6), we use the section formula. The x-axis has the equation y = 0, so the point of division lies on the x-axis, meaning its y-coordinate is 0.

 Step 1: Section formula
The section formula states that if a point (x, y) divides the line segment joining two points (x₁, y₁) and (x₂, y₂) in the ratio m:n, then:

x = (mx₂ + nx₁) / (m + n)
y = (my₂ + ny₁) / (m + n)

Here, the given points are:
(x₁, y₁) = (2, -3)
(x₂, y₂) = (5, 6)

Let the ratio be m:n. Since the point of division lies on the x-axis, its y-coordinate is 0. Using the y-coordinate formula:

y = (my₂ + ny₁) / (m + n)

Substitute y = 0, y₁ = -3, and y₂ = 6:

0 = (m(6) + n(-3)) / (m + n)

Simplify:

0 = (6m – 3n) / (m + n)

Multiply through by (m + n) (which is nonzero):

6m – 3n = 0

Rearrange to solve for the ratio m:n:

6m = 3n
m/n = 3/6
m/n = 1/2

Thus, the ratio is 1:2.

Step 2: Verify the solution
The x-axis divides the line segment in the ratio 1:2. To confirm, substitute m = 1 and n = 2 into the section formula for the y-coordinate:

y = (my₂ + ny₁) / (m + n)
y = (1(6) + 2(-3)) / (1 + 2)
y = (6 – 6) / 3
y = 0

This confirms that the point of division lies on the x-axis.

The correct answer is:
a) 1:2
This question related to Chapter 7 Mathematics Class 10th NCERT. From the Chapter 7 Coordinate Geometry. Give answer according to your understanding.

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