Question

13)Find the coordinates of the point
P(x,y) which divides internally in the
ratio 2:5 the line segment AB, Where
points A(3,-1) and B (10,13). *​

Answer 1

Answer:

Coordinates of P =(5, 3)

Step-by-step explanation:

Given:

• Point A (3 , -1)
• Point B (10, 13)
• Ratio of division = 2 : 5

To Find:

• Point P (x,y)

Concept:

Here we are given the coordinates of the two points of the line segment and the ratio of division. Substituting the value in the section formula and solving it we get the coordinates of the point of division.

Solution:

We have to find the coordiates of point of division.

By section formula we know that,

$\tt{(x,y)=\bigg(\dfrac{m_1x_2+m_2x_1}{m_1+m_2} ,\dfrac{m_1y_2+m_2y_1}{m_1+m_2} \bigg)}$

where m₁ = 2, m₂ = 5, x₁ = 3, x₂ = 10, y₁ = -1, y₂ = 13

Substituting the data we get,

$\tt{(x,y)=\bigg(\dfrac{2\times 10+5\times 3}{2+5} ,\dfrac{2\times 13+5\times -1}{2+5} \bigg)}$

$\tt{(x,y)=\bigg(\dfrac{20+15}{7} ,\dfrac{26+ -5}{7} \bigg)}$

$\tt{(x,y)=\bigg(\dfrac{35}{7} ,\dfrac{21}{7} \bigg)}$

$\tt{(x,y)=(5,3)}$

Hence the coordinates of the point P is (5,3)

Notes:

The section formula is given by,

$\tt{(x,y)=\bigg(\dfrac{m_1x_2+m_2x_1}{m_1+m_2} ,\dfrac{m_1y_2+m_2y_1}{m_1+m_2} \bigg)}$