The point, which divides the line segment joining the points $(7, 6)$ and $(3, 4)$ in ratio $1 : 2$ internally, lies in the

I

quadrant

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I I

quadrant

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I I I

quadrant

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I V

quadrant

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Solution

The correct option is A $I$ quadrant
$L e t t h e l i n e b e A B w i t h p o i n t s A (x_{1}, y_{1}) = (7, 6) & B (x_{2}, y_{2}) = (3, 4) . W e k n o w t h a t t h e c o o r d i n a t e s o f t h e p o i n t, w h i c h d i v i d e s a l i n e s e g m e n t j o i n i n g t h e p o i n t s (x_{1}, y_{1}) & (x_{2}, y_{2}) i n t e r n a l l y i n r a t i o m : n, a r e X = \frac{m x_{2} + n x_{1}}{m + n} a n d Y = \frac{m y_{2} + n y_{1}}{m + n} . ∴ H e r e X = \frac{1 \times 3 + 2 \times 7}{1 + 2} = \frac{17}{3} w h i c h i s p o s i t i v e . A g a i n Y = \frac{1 \times 4 + 6 \times 2}{1 + 2} = \frac{16}{3} w h i c h i s p o s i t i v e . ∵ B o t h t h e c o o r d i n a t e s a r e p o s i t i v e (X, Y) l i e i n t h e f i r s t q u a r d r a n t . A n s - O p t i o n A . N O T E - B o t h t h e p o i n t s a r e i n t h e f i r s t q u a r d r a n t . S o t h e s e g m e n t, j o i n i n g t h e m, l i e s e n t i r e l y i n t h e I q u a r d r a n t . ∴ t h e p o i n t, d i v i d i n g t h e l i n e i n t e r n a l l y, w i l l o b v i o u s l y l i e i n t h e I q u a r d r a n t .$

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