TS EAMCET · Maths · Straight Lines
Two families of lines are given by \(a x+b y+c=0\) and \(4 a^2+9 b^2-c^2-12 a b=0\). Then the line common to both the families is
- A a line passing through \((-1,2)\) and \((2,3)\)
- B a line passing through \((3,2)\) and \((2,3)\)
- C a line passing through \((-3,-2)\) and \((-2,-3)\)
- D a line passing through \((2,-3)\) and \((-2,3)\)
Answer & Solution
Correct Answer
(D) a line passing through \((2,-3)\) and \((-2,3)\)
Step-by-step Solution
Detailed explanation
\((2a-3b)^2 - c^2 = 0\) \((2a-3b-c)(2a-3b+c) = 0\) Substituting \(c = -(ax+by)\) into \((2a-3b-c)(2a-3b+c) = 0\): \((2a-3b+ax+by)(2a-3b-ax-by) = 0\) \((a(2+x)+b(y-3))(a(2-x)+b(-3-y)) = 0\) This means for the point \((x,y)\), either \(2+x=0, y-3=0\) or \(2-x=0, -3-y=0\). The…
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