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

Solution:

Given, the line segment joining the points (-4, -6) and (-1, 7)

We have to find the ratio of division of the line segment and the coordinates of the point of division.

By section formula,

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 k : 1 are [(kx₂ + x₁)/(k + 1) , (ky₂ + y₁)/(k + 1)]

Here, (x₁ , y₁) = (-4, -6) and (x₂ , y₂) = (-1, 7)

So, [(k(-1) + (-4))/(k + 1) , (k(7) + (-6))/(k + 1)] = k:1

[(-k - 4)/(k + 1), (7k - 6)/(k + 1)] = k:1

The point lies on the x-axis. i.e.,y = 0

So, 7k - 6/k + 1 = 0

7k - 6 = 0

7k = 6

k = 6/7

Therefore, the ratio of division is 6:7.

To find the coordinates of the point of division,

x coordinates is (m₁x₂ + m₂x₁)/(m₁ + m₂)

Here, m₁:m₂ = 6:7, (x₁ , y₁) = (-4, -6) and (x₂ , y₂) = (-1, 7)

= [6(-1) + 7(-4)]/(6 + 7)

= -6 - 28/13

= -34/13

Therefore, the coordinate of the point of division is (-34/13, 0).

✦ Try This: In what ratio is the line segment joining A(2, -3) and B(5, 6) divide by the x-axis? Also, find the coordinates of the pint of division.

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 7

NCERT Exemplar Class 10 Maths Exercise 7.3 Problem 10

Summary:

The x–axis divides the line segment joining the points (– 4, – 6) and (–1, 7) in the ratio 6:7. The coordinates of the point of division is (-34/13, 0)

☛ Related Questions:

Question:

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

Solution:

Let AB be divided by the x-axis in the ratio k : 1 at the point P.
Then, by section formula the coordinates of P are

$P=\left(\frac{5 k+2}{k+1}, \frac{6 k-3}{k+1}\right)$

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

Therefore, $\frac{6 k-3}{k+1}=0$

$\Rightarrow 6 k-3=0 \Rightarrow 6 k=3 \Rightarrow k=\frac{3}{6} \Rightarrow k=\frac{1}{2}$

Therefore, the required ratio is $\frac{1}{2}: 1$, which is same as $1: 2$.

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

Applying $k=\frac{1}{2}$, we get the coordinates of point:

$P\left(\frac{5 k+2}{k+1}, 0\right)$

$=P\left(\frac{5 \times \frac{1}{2}+2}{\frac{1}{2}+1}, 0\right)$

$=P\left(\frac{\frac{5+4}{2}}{\frac{1+2}{2}}, 0\right)$

$=P\left(\frac{9}{3}, 0\right)$

$=P(3,0)$

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

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)

Was This helpful?

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).