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

1.	The ratio in which (-6,5) divide the join of (-3,-1) and (-8,9) is?
Answer» $\large{\underline{\bf{\red{Given:-}}}}$ ✦ (-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 :- $\bf\purple{(x,y) = ( \frac{mx_2 + nx_1}{m + <klux>N</klux>}, \frac{my_2 + my_1}{m + n} )}\\$ 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. ━━━━━━━━━━━━━━━━━━━━━━━━━

Answer»

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

✦ (-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 :-

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

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

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