In what ratio is the line joining the points 2 3 and 6 5 divided by the x-axis also find the coordinates of the point of division?

Question 34 Co-ordinate Geometry Exercise 14.3

Answer:

Solution:

The coordinates of the point that splits a line segment (internally or externally) into a specific ratio are found using the Section formula.

Let A(-2, -3) and B(5, 6) be the given points.

Suppose x-axis divides AB in the ratio k: 1 at the point P

Then, the coordinates of the point of division are

\left[\frac{5 k-2}{k+1}, \frac{6 k-3}{k+1}\right]

As, P lies in the x-axis, the y – coordinate is zero.

So,

6k – 3/ k + 1 = 0

6k – 3 = 0

k = ½

Thus, the required ratio is 1: 2

Using k in the coordinates of P

We get, P (1/3, 0)

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Let AB be divided by the x-axis in the ratio k : 1 at the point P.

Then, by section formula the coordination of P are

`p = ((5k+2)/(k+1),(6k-3)/(k+1))`

But P lies on the x-axis; so, its ordinate is 0.

Therefore , `(6k-3)/(k+1) = 0`

`⇒ 6k -3=0 ⇒ 6k =3 ⇒k = 3/6 ⇒ k = 1/2`

Therefore, the required ratio is `1/2:1 `, which is same as 1 : 2

Thus, the x-axis divides the line AB li the ratio 1 : 2 at the point P.

Applying `k=1/2` we get the coordinates of point.

`p((5k+1)/(k+1) , 0)`

`=p((5xx1/2+2)/(1/2+1),0)`

`= p (((5+4)/2)/((5+2)/2),0)`

`= p (9/3,0)`

= p (3,0)

Hence, the point of intersection of AB and the x-axis is P( 3,0).

Page 2

Let AB be divided by the x-axis in the ratio :1 k at the point P.

Then, by section formula the coordination of P are

`p = ((3k-2)/(k+1) , (7k-3)/(k+1))`

But P lies on the y-axis; so, its abscissa is 0.
Therefore , `(3k-2)/(k+1) = 0`

`⇒ 3k-2 = 0 ⇒3k=2 ⇒ k = 2/3 ⇒ k = 2/3 `

Therefore, the required ratio is `2/3:1`which is same as 2 : 3
Thus, the x-axis divides the line AB in the ratio 2 : 3 at the point P.

Applying `k= 2/3,`  we get the coordinates of point.

`p (0,(7k-3)/(k+1))`

`= p(0, (7xx2/3-3)/(2/3+1))`

`= p(0, ((14-9)/3)/((2+3)/3))`

`= p (0,5/5)`

= p(0,1)

Hence, the point of intersection of AB and the x-axis is P (0,1).