Question

The ratio in which the line segment joining the points PC-3, 10) and Q(6,- S) is divided by
Ol-1, 6) is

Answer 1

Answer:

The coordinates of the point P(x, y) which divides the line segment joining the points A(x₁, y₁) and B(x₂, y₂), internally, in the ratio m₁: m₂ is given by the Section Formula: P(x, y) = [(mx₂ + nx₁) / m + n, (my₂ + ny₁) / m + n]

Let the ratio in which the line segment joining j

7k = 2

k = 2 / 7

Hence, the point C divides line segment AB in the ratio 2 : 7.

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Answer 2

Given,

$\[P\left( { - 3,10} \right),Q\left( {6, - 8} \right)\]$ divided by $\[O\left( { - 1,6} \right)\]$

Solution,

The coordinates of the point $\[P\left( {x,y} \right)\]$ which divides the line segment joining the points$\[A\left( {{x_1},{\rm{ }}{y_1}} \right){\rm{ }}and{\rm{ }}B\left( {{x_2},{\rm{ }}{y_2}} \right),\]$in the ratio of $\[m:n\]$ is given by the Section Formula,

$\[P\left( {x,y} \right) = \left\{ {\left( {\frac{{m{x_2} + n{x_1}}}{{m + n}}} \right),\left( {\frac{{m{y_2} + n{y_1}}}{{m + n}}} \right)} \right\}\]$

$\[O\left( { - 1,6} \right) = \left\{ {\left( {\frac{{m \times 6 + n\left( { - 3} \right)}}{{m + n}}} \right),\left( {\frac{{m \times \left( { - 8} \right) + n \times 10}}{{m + n}}} \right)} \right\}\]$

Compare the both coordinate,

$\[\begin{array}{l}6m - 3n = - m - n\\ \Rightarrow 7m - 2n = 0\\ \Rightarrow 7m = 2n - - - - \left( 1 \right)\\ - 8m + 10n = 6m + 6n\\ \Rightarrow - 14m + 4n = 0\\ \Rightarrow 7m = 2n - - - - \left( 2 \right)\end{array}\]$

Consider,

$\[\begin{array}{l}m = 2\\n = 7\end{array}\]$

Hence the ratio is $\[2:7\]$