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

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

In what ratio is the line segment joining a 2 and B 5/6 divided by the x-axis also find the coordinates of the point of division?
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Question 5 Coordinated Geometry - Exercise 7.3

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

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)

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

In what ratio does the x axis divide the join of A2, 3 and B 5, 6? a 2 : 3 b 3 : 5 c 1 : 2 d 2 : 1

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6k3k+1=0


6k3=0


k=36=12

Thus, x-axis divides the line segment joining the points (2, –3) and (5,6) in the ratio 1:2.


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