Math, asked by babariyals123, 1 month ago

Find the coordinates of the point which divides the line segment joining points A (-3,4)

and B (13,12) in the ratio 5:3​

Answers

Answered by Ladylaurel
13

Answer :-

The required co-ordinates is (7,9).

Step-by-step explanation:

We know,

 \bf{ \underline{Co-ordinate \: of \: P}} \\ \bigg \{ \sf{\dfrac{{m}_{1}{x}_{2} + {m}_{2}{x}_{1}}{{m}_{1} + {m}_{2}}} \: , \: \sf{\dfrac{{m}_{1}{y}_{2} + {m}_{2}{y}_{1}}{{m}_{1} + {m}_{2}}} \bigg \}

⠀⠀⠀⠀⠀ ____________________

Let the co-ordinate of the point be P(x,y).

\sf{\therefore \:  \:  \: x = \sf{\dfrac{{m}_{1}{x}_{2} + {m}_{2}{x}_{1}}{{m}_{1} + {m}_{2}}}}

\sf{= \sf{\dfrac{{m}_{1}{x}_{2} + {m}_{2}{x}_{1}}{{m}_{1} + {m}_{2}}}}

\sf{= \sf{\dfrac{5 \times 13 + 3 \times  - 3}{5 + 3}}}

\sf{= \sf{\dfrac{65 + ( - 9)}{5 + 3}}}

\sf{= \sf{\dfrac{65 - 9}{5 + 3}}}

\sf{= \sf{\dfrac{56}{5 + 3}}}

\sf{= \sf{\dfrac{56}{8}}}

\sf{= \red{7}}

and, \sf{y = \sf{\dfrac{{m}_{1}{y}_{2} + {m}_{2}{y}_{1}}{{m}_{1} + {m}_{2}}}}

 \sf{ = \dfrac{{m}_{1}{y}_{2} + {m}_{2}{y}_{1}}{{m}_{1} + {m}_{2}}}

 \sf{ = \dfrac{5 \times 12 + 3 \times 4}{5 + 3}}

 \sf{ = \dfrac{60 + 12}{5 + 3}}

 \sf{ = \dfrac{72}{5 + 3}}

 \sf{ = \dfrac{72}{8}}

 \sf{ = \red{9}}

Therefore, \underline{\boxed{\textsf{\textbf{P = \{ 7,9 \}}}}}

Attachments:
Answered by ItzWhiteStorm
28

Solution:-

  • Here,The coordinates of the point which divides the line segment joining A (- 3,4) and B (13,12) in the ratio 5:3.Let the us apply the section formulae of internally where m + n ≠ 0.

 \\  \large\underline{\bigstar \: \bf{Formula \:  \bigstar}} \\   \\    \sf{\overline{AB} =   \bigg(\frac{mx_2+nx_1}{m + n}  \frac{,my_2+ny_1}{,m + n}  \bigg)} \\  \\

Let A(x1,y1) = (-3,4) and B(x2,y2) = (13,12).

Given ratio m : n = 5:3

\\ \\

Putting the values on formula,

 \\  \\ \dashrightarrow \sf{ \bigg(  \frac{(5  \times 13)  + (3  \times ( - 3))}{5 + 3}, \frac{(5 \times 12) +(3  \times 4) }{5 + 3}  \bigg)} \\   \\ \dashrightarrow \sf{ \bigg( \frac{65 + ( - 9)}{8} , \frac{60 + 12}{8}  \bigg)} \\  \\ \dashrightarrow \sf{ \bigg( \frac{56}{8},  \frac{72}{8}  \bigg)} \\  \\ \dashrightarrow \sf{7,9}\\ \\

Hence,

  • The coordinate points are (7,9).

_______________________________

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