Question

find the ratio in which the point (2, 5) divides the line segment (8,2) and(-6,9)​

Answer 1

The required ratio is 3:4.

Step-by-step explanation:

Since we have given that

(2,5)  divides the line segment (8,2) and (-6,9)

Let the ratio be k:1.

Using "Section formula", we get that

$(\dfrac{8-6k}{k+1},\dfrac{9-2k}{k+1})=(2,5)\\\\\dfrac{8-6k}{k+1}=2\\\\8-6k=2(k+1)\\\\8-6k=2k+2\\\\8-2=2k+6k\\\\6=8k\\\\k=\dfrac{6}{8}=\dfrac{3}{4}$

Hence, the required ratio is 3:4.

# learn more:

Find the coordinates of the point dividing internally the line segment joining a(3,7) and b(8,2) in the ratio 2:3.

https://brainly.in/question/8281975

Answer 2

$Let \: P(2,5) = (x,y) \:divide \:the \\line \:segment \:A(8,2) = (x_{1},y_{1}) \\and \:B(-6,9) = (x_{2},y_{2})\: internally \\ in \:the \: ratio \:m : n.$

$\underline { \blue{By \:Section \: Formula :}}$

$( x, y ) = \big( \frac{mx_{2}+nx_{1}}{m+n} , \frac{my_{2}+ny_{1}}{m+n}\big)$

$\implies ( 2,5) = \big( \frac{m(-6)+n\times 8}{m+n} , \frac{m\times 9+n\times 8}{m+n}\big)$

$\implies 2 = \frac{-6m+ 8n }{m+n}$

$\implies 2(m+n) = -6m + 8n$

$\implies 2m + 2n = - 6m + 8n$

$\implies 2m + 6m = 8n - 2n$

$\implies 8m = 6n$

$\implies \frac{m}{n} = \frac{6}{8}$

$\implies \frac{m}{n} = \frac{3}{4}$

Therefore.,

$\red { Required \:ratio } \green { = 3 : 4 }$

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