The line segment joining the points A (3, 2) and B (5,1) is divided at the point P in the ratio 1:2 and it lies on the line 3x – 18y + k = 0. Find the value of k
Solution:
Given, the line segment joining the points A(3, 2) and B(5, 1) is divided at the point P in the ratio 1:2
The point P lies on the line 3x - 18y + k = 0
We have to find the value of k.
The coordinates of the point P which divides the line segment joining the points A (x₁ , y₁) and B (x₂ , y₂) internally in the ratio m₁ : m₂ are
[(m₁x₂+m₂x₁)/(m₁+m₂) , (m₁y₂+m₂y₁)/(m₁+m₂)]
Here, m₁:m₂ = 1:2, (x₁ , y₁) = (3, 2) and (x₂ , y₂) = (5, 1)
Let the point P be (x, y)
[(1(5) + 2(3))/(1 + 2), (1(1) + 2(2))/(1 + 2)] = (x, y)
[(5 + 6)/3, (1 + 4)/3] = (x, y)
[11/3, 5/3] = (x, y)
The point P is (11/3, 5/3)
The point P lies on the line 3x - 18y + k = 0
So, 3(11/3) - 18(5/3) + k = 0
11 - 6(5) + k = 0
11 - 30 + k = 0
-19 + k = 0
k = 19
Therefore, the value of k is 19.
✦ Try This: The point P which divides the line segment joining the points A(2,-5) and B(5, 2) in the ratio 2:3 lies in the quadrant
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 7
NCERT Exemplar Class 10 Maths Exercise 7.3 Problem 15
The line segment joining the points A (3, 2) and B (5,1) is divided at the point P in the ratio 1:2 and it lies on the line 3x – 18y + k = 0. Find the value of k
Summary:
The line segment joining the points A (3, 2) and B (5,1) is divided at the point P in the ratio 1:2 and it lies on the line 3x – 18y + k = 0. The value of k is 19
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