Question

The ratio in which (-6,5) divide the join of (-3,-1) and (-8,9) is?

Answer 1

$\large{\underline{\bf{\red{Given:-}}}}$

• ✦ (-6 ,5) divides the line segment joining (-3,-1) and (-8,9)

$\large{\underline{\bf{\red{To\:Find:-}}}}$

• ratio in which (-6,5) divides the line segment.

$\huge{\underline{\bf{\green{Solution:-}}}}$

Let the ratio be k:1

Let coordinates of p are (-6,5)

By section formula :-

$\bf\purple{(x,y) = ( \frac{mx_2 + nx_1}{m + n}, \frac{my_2 + my_1}{m + n} )}$

Let m = k and n = 1

The coordinates of p are

$\mapsto \rm\:p( \frac{k \times - 8 + 1 \times - 3}{k + 1}, \frac{k \times 9 + 1 \times - 1}{k + 1} )\: \\ \\ \mapsto \rm\:p( \frac{ - 8k - 3}{k + 1}, \frac{9k - 1}{k + 1}) \\ \\ \rm\:it \: is \: given \: that \: \: p( - 6,5) \\ so \\ \\ \mapsto \rm\small\frac{ - 8k - 3}{k + 1} = - 6 \: \: and \: \: \frac{9k - 1}{k + 1} = 5 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underbrace{ \bf\:case \: 1st} \\ \\ \mapsto \rm\: \frac{ - 8k - 3}{k + 1} = - 6 \\ \\ \mapsto \rm\: - 8k - 3 = - 6(k + 1) \\ \\ \mapsto \rm\: - 8k - 3 = - 6k - 6 \\ \\ \mapsto \rm\: - 3 + 6 = - 6k + 8k \\ \\\mapsto \rm\:3 = 2k \\ \\\mapsto \bf \blue{\:k = \frac{3}{2} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underbrace {\bf\:2nd \: \: case} \\ \\ \mapsto \rm\: \frac{9k - 1}{k + 1} = 5 \\ \\ \mapsto \rm\:9k - 1 = 5(k + 1) \\ \\\mapsto \rm\:9k - 1 = 5k + 5 \\ \\\mapsto \rm\:9k - 5k = 5 + 1 \\ \\ \mapsto \rm\:4k = 6 \\ \\\mapsto \rm\:k = \cancel\frac{6}{4} \\ \\ \mapsto \bf\blue{k = \frac{3}{2}}$

K = 3/2 in both cases.

So,

The required ratio is 3/2:1

which is 3:1.

Hence ,

(-6 ,5) divides the line segment joining (-3,-1) and (-8,9) in the ratio of 3:2.

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Answer 2

$\huge\purple{\underline{\underline{\pink{Ans}\red{wer:-}}}}$

$\sf{The \ point \ divides \ the \ line \ in \ ratio}$

$\sf{of \ 3:2}$

$\sf\orange{Given:}$

$\sf{\implies{Point(-6,5) \ divides \ the \ line \ joining}}$

$\sf{points \ (-3,-1) \ and \ (-8,9)}$

$\sf\pink{To \ find:}$

$\sf{The \ ratio \ in \ which \ line \ is \ divided.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{Let,}$

$\sf{x1=-3,y1=1,x2=-8,y2=9,x=-6,y=5}$

$\sf{By \ section \ formula}$

$\sf{x=\frac{mx2+nx1}{m+n} \ ;y=\frac{my2+ny1}{m+n}}$

$\sf{\implies{-6=\frac{m(-8)+n(-3)}{m+n}}}$

$\sf{\implies{-6(m+n)=-8m+3n}}$

$\sf{\implies{-6m-6n=-8m-3n}}$

$\sf{\implies{-6m+8m=-3n+6n}}$

$\sf{\implies{2m=3n}}$

$\sf{\implies{\frac{m}{n}=\frac{3}{2}}}$

$\sf\purple{\tt{\therefore{The \ point \ divides \ the \ line \ in \ ratio}}}$

$\sf\purple{\tt{of \ 3:2}}$