Question

Find the co - ordinates of the point P which divides the line segment joining A (8 , 9) and B (-7 , 4) internally in the ratio 2 : 3.

Answer 1

$\Large{\bf{\red{\mathfrak{\dag{\underline{\underline{Given:-}}}}}}}$

• A (8,9) and B (- 7,4) are divided internally in the ratio 2:3.

$\Large{\bf{\orange{\mathfrak{\dag{\underline{\underline{To \: Find:-}}}}}}}$

• Co - ordinates of point P (x, y).

$\Large{\bf{\purple{\mathfrak{\dag{\underline{\underline{Solution:-}}}}}}}$

• A (8,9) and B (- 7,4) are divided internally in the ratio 2:3.

• Co - ordinates of point P (x, y) =

• We know that,

$\boxed{\pink{\sf Section \: Formula \: = \: \dfrac{m_{1}x_{2} \: + \: m_{2}x_{1}}{m_{1} \: + \: m_{2}} \: , \: \dfrac{m_{1}y_{2} \: + \: m_{2}y_{1}}{m_{1} \: + \: m_{2}}}}$

• From, A (8,9) and B (- 7,4)

• Here,

• $\sf x_{1} \: = \: 8$

• $\sf x_{2} \: = \: - \: 7$

• $\sf y_{1} \: = \: 9$

• $\sf y_{2} \: = \: 4$

• $\sf m_{1} : m_{2} \: = \: 2 : 3$

• Substituting the values,

$\sf P(x,y) \: = \: \dfrac{2 \: * \: - \: 7 \: + \: 3 \: * \: 8}{2 \: + \: 3} \: , \: \dfrac{2 \: * \: 4 \: + \: 3 \: * \: 9}{2 \: + \: 3}$

$\sf P(x,y) \: = \: \dfrac{ - \: 14 \: + \: 24}{5} \: , \: \dfrac{8 \: + \: 27}{5}$

$\sf P(x,y) \: = \: \dfrac{10}{5} \: , \: \dfrac{35}{5}$

$\sf P(x,y) \: = \: \dfrac{\cancel{10}}{\cancel{5}} \: , \: \dfrac{\cancel{35}}{\cancel{5}}$

$\sf P(x,y) \: = \: \dfrac{2}{1} \: , \: \dfrac{7}{1}$

$\sf P(x,y) \: = \: 2 \: , \: 7$

Therefore,

• $\underline{\boxed{\blue{\textsf{Co - ordintes of P (x, y) = P (2, 7).}}}}$

$\Large{\bf{\red{\mathfrak{\dag{\underline{\underline{Formulas \: Used:-}}}}}}}$

$\sf Section \: Formula \: = \: \dfrac{m_{1}x_{2} \: + \: m_{2}x_{1}}{m_{1} \: + \: m_{2}} \: , \: \dfrac{m_{1}y_{2} \: + \: m_{2}y_{1}}{m_{1} \: + \: m_{2}}$

Answer 2

Answer:

Given :-

A(8,9)

B(-7,4)

Ratio = 2:3

To Find :-

Co-ordinate of point P

Solution :-

By using the section formula

$\dag{\boxed{\red{\mathfrak{\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2},\dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}}}}}$

We have

$\begin{cases} \sf x_1 = 8 \\ \sf \: x_2 = - 7 \\ \sf \: y_1 = 9 \\ \sf \: y_2 = 4 \\ \sf m_1 = 2 \\ \sf \: m_2 = 3\end{cases}$

By putting the value

$\sf \pink{\dfrac{2 \times - 7 + 3 \times 8}{2 + 3}, \dfrac{2 \times 4 + 3 \times 9}{2 + 3}}$

$\sf \blue{ \dfrac{ - 14 + 24}{2 + 3} , \dfrac{8 + 27}{2 + 3}}$

$\sf \red { \dfrac{10}{2 + 3} , \dfrac{35}{2 + 3}}$

$\sf \pink{ \dfrac{10}{5} , \dfrac{35}{5} }$

$\sf \red{ (2,7)}$