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

Question 5 Coordinated Geometry - Exercise 7.3

Answer:

Let the ratio in which x-axis divides the line segment joining (–4, –6) and (–1, 7) = 1: k.

Then,

x-coordinate becomes, $\frac{\left(-1-4k\right)}{(k+1)}$

y-coordinate becomes, $\frac{\left(7-6k\right)}{(k+1)}$

Since P lies on x-axis, y coordinate = 0

$\frac{\left(7-6k\right)}{(k+1)}=0\\ 7-6k=0\\ k=\frac{7}{6}$

Therefore, the point of division divides the line segment in the ratio 6 : 7.

Now, m1 = 6 and m2 = 7

By using the section formula,

$x=\frac{\left(m_1x_2+m_2x_2\right)}{(m_1+m_2)}=\frac{\left[6(-1)+7(-4)\right]}{(6+7)}=\frac{\left(-6-28\right)}{13}=-\frac{34}{13}\\ So,\ now\\ y=\frac{\left[6(7)+7(-6)\right]}{(6+7)}=\frac{\left(42-42\right)}{13}=0$

Hence, the coordinates of P are (-34/13, 0)

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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:

Text Solution

Answer : `(1:2), (3, 0)`

Solution : Let the x-axis be cut by the join of A(2, -3) and B(5, 6) in the ratio k:1 at the point P(x, 0). <br> Then, `(6k-3)/(k+1) = 0 rArr 6k - 3 =0 rArr 6k = 3 rArr k = (3)/(6) = (1)/(2). ` <br> `therefore "required ratio is" (1)/(2): 1, i.e., 1:2.` <br> Point of division is `P((1 xx 5 + 2 xx 2)/(1+2), 0), i.e., (3, 0).`

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)`