| List - I | List - II | ||
| P. | Distance between (- 6, 7) and (-1, -5) is | 1. | - 3, 7 |
| Q. | The value of k for which the distance between \[A\left( k,\,-5 \right)\] and \[B\left( 2,\,7 \right)\] is 13 units | 2. | x + y = 5 |
| R. | (x, y) is equidistant from (5, 1) and (-1, 5) if | 3. | 3x = 2y |
| S. | (x, y), (2, 3) and (4, 1) are collinear, if | 4. | 13 units |
A) P-3, Q-4, R-2, S-1
B) P-1, Q-2, R-4, S-3
C) P-4, Q-1, R-3, S-2
D) P-3, Q-2, R-4, S-1
Correct Answer: C
Solution :
| (P) Distance between (- 6,7) and (-1, - 5) |
| \[=\sqrt{{{\left( -1+6 \right)}^{2}}+{{\left( -5-7 \right)}^{2}}}=\sqrt{{{\left( 5 \right)}^{2}}+{{\left( 12 \right)}^{2}}}\] |
| \[=\sqrt{25+144}=\sqrt{169}=13\,\]units |
| (Q) The given points are A (k, - 5) and B (2, 7). |
| Now, \[AB=13\Rightarrow A{{B}^{2}}=169\] |
| \[\Rightarrow \,\,{{\left( 2-k \right)}^{2}}+{{\left( 7+5 \right)}^{2}}=169\] |
| \[\Rightarrow \,\,{{k}^{2}}-4k+4+144=169\] |
| \[\Rightarrow \,\,{{k}^{2}}-4k-21=0\] |
| \[\Rightarrow \left( k-7 \right)\left( k+3 \right)=0\] |
| \[\Rightarrow \,\,k=7\] or \[k=-3\] |
| [R] P (x, y), A(5, 1), B(-1, 5) |
| \[PA=PB\] |
| \[\Rightarrow \,\,\,P{{A}^{2}}=P{{B}^{2}}\] |
| \[\Rightarrow \,\,{{\left( 5-x \right)}^{2}}+{{\left( 1-y \right)}^{2}}={{\left( -1-x \right)}^{2}}+{{\left( 5-y \right)}^{2}}\] |
| \[\Rightarrow \,\,25+{{x}^{2}}-10x+1+{{y}^{2}}-2y\] |
| \[=1+{{x}^{2}}+2x+25+{{y}^{2}}-10y\] |
| \[\Rightarrow \,\,3x=2y\] |
| (S) (x, y), (2, 3) and (4, 1) are collinear, then x(3 - 1) + 2(1 - y) + 4 (y - 3) = 0 |
| \[\Rightarrow \,\,\,2x+2-2y+4y-12=0\] |
| \[\Rightarrow \,\,\,x+y=5\] |
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