FIND THE RATIO THAT X AXIS DIVIDES THE LINE SEGMENT JOINING THE POINTS

The coordinates of the point P(x, y) which divides the line segment joining the points A (x₁, y₁) and B (x₂, y₂) internally in the ratio l : m

Problem 1 :

Find the ratio in which the line segment joining A(1, -5) and B(-4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

Solution:

Given, A(1, -5) and B(-4, 5)

Let the x axis divide the line segment at point (x, 0) in the ratio k : 1.

Using section formula,

So, the required ratio is 1 : 1.

Here, the point of division is P(x, 0).

On comparing x coordinate,

Hence, the required ratio is 1 : 1 and the coordinates of the point of division is (-3/2, 0).

Problem 2 :

In what ratio is the line segment joining A(6, 3) and B(-2, -5) is divided by the x-axis. Also find the coordinates of the point of intersection of AB and the x-axis.

Solution:

Using the section formula, if a point (x, y) divides the line joining the points (x₁, y₁) and (x₂, y₂) in the ratio of l : m, then

Let point P on the x axis divides the line segment joining the points A and B the ratio l : m

Consider P lies on x axis having coordinates (x, 0).

Then,

3(x + 2)=5(6 - x)

3x + 6 = 30 - 5x

8x = 24

x = 3

Hence, the coordinates are (3, 0).

Problem 3 :

In what ratio is the line segment joining A(2, -3) and B(5, 6) is divided by the x-axis. Also find the coordinates of the point of intersection of AB and the x-axis.

Solution:

Using the section formula, if a point (x, y) divides the line joining the points (x₁, y₁) and (x₂, y₂) in the ratio of l : m, then

Let point P on the x axis divides the line segment joining the points A and B the ratio l : m

Consider P lies on x axis having coordinates (x, 0).

l : m = 1 : 2

Then,

(x - 5)(1) = (2 - x)(2)

x - 5 = 4 - 2x

3x = 9

x = 3

Hence, the coordinates are (3, 0).

Problem 4 :

Find the point P on the x-axis which is equidistant from the points A(5, 4) and B(-2, 3). Also find the area of ΔPAB.

Solution:

Let the point of x axis be P(x, 0) = (x₁, y₁)

Given, A(5, 4) and B(-2, 3) are equidistant from P.

PA = PB

PA² = PB² --> (1)

Distance between two points is

From equation (1)

x² - 10x + 41 = x² + 4x + 13

-10x - 4x = 13 - 41

-14x = -28

x = 2

Therefore, coordinates of P is (2, 0).

Now, Area of  ΔPAB

Hence, point P is (2,0) and area of ΔPAB is 12.5 square unit.