List - I | List - II | ||
P. | A(-1, 3) and B (-5, 6) internally in the ratio 1 : 2 | 1. | (7, 3) |
Q. | A (-2, 1) and B (1, 4) internally in the ratio 2 : 1 | 2. | (0, 3) |
R. | A (1, 7) and B (3, 4) internally in the ratio 3 : 3 | 3. | \[\left( \frac{11}{5},\frac{26}{5} \right)\] |
S. | A (4, -3) and B (8, 5) internally in the ratio 3 : 1 | 4. | \[\left( -\frac{7}{3},\,4 \right)\] |
A) P-4, Q-2, R-3, S-1
B) P-3, Q-2, R-4, S-1
C) P-1, Q-4, R-3, S-2
D) P-3, Q-1, R-2, S-4
Correct Answer: A
Solution :
(P) |
\[x=\frac{-5-2}{3}=-\frac{7}{3}\Rightarrow \,\,y=\frac{6+6}{3}=4\] |
\[\therefore\] Point P is \[\left( \frac{-7}{3},\,4 \right)\] |
(Q) |
\[x=\frac{2-2}{2+1}=0,\,y=\frac{8+1}{2+1}=3\] |
\[\therefore\] Points P is (0, 3) |
[R] |
\[x=\frac{9+2}{5}=\frac{11}{5},\,y=\frac{12+14}{5}=\frac{26}{5}\] |
\[\therefore\] Points P is \[\left( \frac{11}{5},\,\frac{26}{5} \right)\] |
(S) |
\[x=\frac{24+4}{3+1}=7,\,y=\frac{15-3}{3+1}=3\] |
\[\therefore\] Point P is (7, 3) |
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