| Directions: Each of these questions contains two statements: Assertion [A] and Reason [R]. Each of these questions also has four alternative choices, any one of which is the correct answer. You have to select one of the codes [a], [b], [c] and [d] given below. |
| Assertion Points (3, 2), (-2, - 3) and (2, 3) form a right triangle. |
| Reason If (x, y) is equidistant from (3, 6) and (-3, 4), then 3x + y = 5. |
A) A is true, R is true; R is a correct explanation for A.
B) A is true, R is true; R is not a correct explanation for A.
C) A is true; R is False.
D) A is false; R is true.
Correct Answer: B
Solution :
| Let \[A\left( 3,\,2 \right),\,B\left( -2,\,-3 \right)\] and\[C\left( 2,\,3 \right)\]. |
| \[\therefore \,\,AB=\sqrt{{{\left( -2-3 \right)}^{2}}+{{\left( -3-2 \right)}^{2}}}=\sqrt{50}\]units |
| \[BC=\sqrt{{{\left( -2-2 \right)}^{2}}+{{\left( -3-3 \right)}^{2}}}=\sqrt{52}\]units |
| and units \[CA=\sqrt{{{\left( 2-3 \right)}^{2}}+{{\left( 3-2 \right)}^{2}}}=\sqrt{2}\] units |
| \[\therefore \,B{{C}^{2}}=A{{B}^{2}}+C{{A}^{2}}\] |
| \[\Rightarrow \Delta ABC\]is a right triangle. |
| Let A'(3,6), 5'(-3,4) and P (x, y) |
| Since, P is equidistant from A' and B', then |
| PA' = PB' |
| \[\Rightarrow PA{{'}^{2}}=PB{{'}^{2}}\] |
| \[\Rightarrow \,\,{{\left( x-3 \right)}^{2}}+{{\left( y-6 \right)}^{2}}={{\left( x+3 \right)}^{2}}+{{\left( y-4 \right)}^{2}}\] |
| \[\Rightarrow \,\,{{x}^{2}}-6x+9+{{y}^{2}}-12y+36\] |
| \[={{x}^{2}}+6x+9+{{y}^{2}}-8y+16\] |
| \[\Rightarrow \,\,12x+4y=20\] |
| \[\Rightarrow \,\,3x+y=5\] |
| Both the statements are true but the Reason is not correct explanation of the Assertion. |
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