List - II gives the coordinates of a point P that divides the line segment AB joining the points given in List -I- Match them correctly- lbeginarray - Wline-List-I- -List IINQ - A -2-1 - and - B 1-4 internally in the ratio 2- 1 -2 -0-3 ll hline e n d array A- P- 4- Q -2- R -3- S -1B- P -3- Q -2- R -4- S -1C- P- 1- Q -4- R -3- S -2D- P -3- Q -1- R -2- S -4

The correct option is A P 4, Q 2, R 3, S 1 (P) x= 5 2/3= 7/3 y=6+6/3=4 ∴ Point is ( 7/3,4 ) (Q) x=2 2/2+1=0, y=8+1/2+1=3 ∴ Point is (0,3) (R) x=9+ ...

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Byju's Answer

Standard IX

Mathematics

Section Formula

List - II giv...

Question

List - II gives the coordinates of a point P that divides the line segment AB joining the points given in List -I. Match them correctly.
$\begin{matrix} L i s t - I & L i s t I I (P) & A (- 1, 3) a n d B (- 5, 6) I n t e r n a l l y i n t h e r a t i o 1 : 2 & (1) & (7, 3) (Q) & A (- 2, 1) a n d B (1, 4) i n t e r n a l l y i n t h e r a t i o 2 : 1 & (2) & (0, 3) (R) & A (1, 7) a n d B (3, 4) i n t e r n a l l y i n t h e r a t i o 3 : 2 & (3) & (\frac{11}{5}, \frac{26}{5}) (S) & a (4, - 3) a n d B (8, 5) i n t e r n a l l y i n t h e r a t i o 3 : 1 & (4) & (- \frac{7}{3}, 4) \end{matrix}$

P- 4, Q-2, R-3, S-1

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P- 3, Q-2, R-4, S-1

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P- 1, Q-4, R-3, S-2

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P- 3, Q-1, R-2, S-4

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Solution

The correct option is A
P- 4, Q-2, R-3, S-1

(P)

$x = \frac{- 5 - 2}{3} = - \frac{7}{3} y = \frac{6 + 6}{3} = 4$
$∴$ Point is $(\frac{- 7}{3}, 4)$

(Q)

$x = \frac{2 - 2}{2 + 1} = 0, y = \frac{8 + 1}{2 + 1} = 3$
$∴$ Point is (0,3)

(R)

$x = \frac{9 + 2}{5} = \frac{11}{5}, y = \frac{12 + 14}{5} = \frac{26}{5}$
$∴$ Point is $(\frac{11}{5}, \frac{26}{5})$

(S)

$x = \frac{24 + 4}{3 + 1} = 7$ ,
$y = \frac{15 - 3}{3 + 1} = 3$
$∴$ Point is (7,3)

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Similar questions

List-II gives distance between pair of points given in List-I. Match them correctly.
$\begin{matrix} L i s t - I & L i s t I I (P) & (- 5, 7), (- 1, 3) & (1) & 5 (Q) & (5, 6), (1, 3) & (2) & \sqrt{8} (R) & (\sqrt{3} + 1, 1), (0, \sqrt{3}) & (3) & \sqrt{6} (S) & (0, 0), (- \sqrt{3}, \sqrt{3}) & (4) & 4 \sqrt{2} \end{matrix}$

Q. List-II gives distance between pair of points given in List-I, match them correctly.

\begin{matrix} L i s t - I & L i s t I I (P) & (- 5, 7), (- 1, 3) & (1) & 5 (Q) & (5, 6), (1, 3) & (2) & \sqrt{8} (R) & (\sqrt{3} + 1, 1), (0, \sqrt{3}) & (3) & \sqrt{6} (S) & (0, 0), (- \sqrt{3}, \sqrt{3}) & (4) & 4 \sqrt{2} \end{matrix}

$\begin{matrix} L i s t - I & L i s t - I I P . & I f_{α} = \frac{π}{7} t h e n \frac{1}{c o s α} + \frac{2 c o s α}{c o s 2 α} = & 1. & 2 Q . & u n d e r s e t x \to \infty L t [(x - 1) (x - 2) (x + 3) (x + 5) (x + 10)]^{\frac{1}{5}} - x = & 2. & 3 R . & \int_{- 2}^{2} \frac{3 x^{2} d x}{1 + e^{x}} = & 3. & 4 S . & L e t f (x) = x^{3} + x^{2} + 2 x - 1. T h e m i n i m u m i n t e g r a l v a l u e o f x i f x s a t i s f i e s f (f (x)) > f (2 x + 1) i s & 4. & 8 \end{matrix}$

List-II gives distance between pair of points given in List-I, match them correctly.
$\begin{matrix} L i s t - I & L i s t I I (P) & (- 5, 7), (- 1, 3) & (1) & 5 (Q) & (5, 6), (1, 3) & (2) & \sqrt{8} (R) & (\sqrt{3} + 1, 1), (0, \sqrt{3}) & (3) & \sqrt{6} (S) & (0, 0), (- \sqrt{3}, \sqrt{3}) & (4) & 4 \sqrt{2} \end{matrix}$

Q. Match the electronic configurations of the elements given in List - I with their correct characteristic(s) (i.e., properties for given configuration) given in List - II.

\begin{matrix} List-I & List - II P. & 1 s^{2} & 1. & Element shows highest negative oxidation state Q. & 1 s^{2} 2 s^{2} 2 p^{5} & 2. & Element shows highest first ionisation enthalpy R. & 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{5} & 3. & Element shows highest electronegativity on Pauling scale S. & 1 s^{2} 2 s^{2} 2 p^{2} & 4. & Element shows maximum electron gain enthalpy (most exothermic) \end{matrix}

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The correct option is A P- 4, Q-2, R-3, S-1 (P) x=−5−23=−73y=6+63=4 ∴ Point is (−73,4) (Q) x=2−22+1=0,y=8+12+1=3 ∴ Point is (0,3) (R) x=9+25=115,y=12+145=265 ∴ Point is (115,265) (S) x=24+43+1=7, y=15−33+1=3 ∴ Point is (7,3)