Question

Find the co-ordinates of the point which divides the line segment joining the points (4, -3) and (8,5) are in the ratio 3 : 1 internally.

Answer 1

Answer:

• (7, 3) is the required point.

Step-by-step explanation:

Let p(x , y) be the required point.

We have:

• A(4 , -3)
• B(8 , 5)
• m₁ : m₂ = 3 : 1

Using section formula:

• ${\dfrac{m_{1}x_{2}+m_{2}x_{1}}{m_1+m_2} ;\dfrac{m_{1}y_{2}+m_{2}y_{1}}{m_1+m_2} }$

Where:

• $\bold{x_1=4}$

• $\bold{x_2=8}$

• $\bold{y_1=-3}$

• $\bold{y_2=5}$

• $\bold{m_1=3}$

• $\bold{m_2=1}$

Substituting we get:

• $\bold{\dfrac{3(8)+1(4)}{3+1} =\dfrac{28}{4}=7 }$ - (x)
• $\bold{\dfrac{3(5)+1(-3)}{3+1}\dfrac{15-3}{4}=3 }$ - (y)

∴ (x , y) = (7 , 3) is the required point.

Answer 2

Given :-

Points (4, -3) and (8,5)

Ratio = 3 : 1

To Find :-

Co-ordinates

Solution :-

We know that

$\sf Section\;formula=(x,y)=\bigg\{\dfrac{mx_2+nx_1}{m+n},\dfrac{mx_1+nx_2}{m+n}\bigg\}$

$\sf (x,y)=\bigg\{\dfrac{3\times8+1\times4}{1+3},\dfrac{3\times5+1\times (-3)}{1+3}\bigg\}$

$\sf (x,y)=\bigg\{\dfrac{24+4}{4},\dfrac{15+(-3)}{4}\bigg\}$

$\sf (x,y)=\bigg\{\dfrac{28}{4},\dfrac{15-3}{4}\bigg\}$

$\sf (x,y)=\bigg\{\dfrac{28}{4},\dfrac{12}{4}\bigg\}$

$\sf (x,y)=(7,3)$