AP EAMCET · Maths · Straight Lines
The locus of a point which moves such that the area of the triangle formed by it with the vertices \((1,2)\) and \((-2,5)\) is 8 sq. units is/are
- A \(3 x+3 y+7=0 \quad \& \quad x+y+3=0\)
- B \(3 x+3 y-25=0 \& \quad x+y+3=0\)
- C \(3 x+3 y-2=0 \quad \& \quad 3 x+3 y-25=0\)
- D \[
3 x+3 y+7=0 \quad \& \quad 3 x+3 y-25=0
\]
Answer & Solution
Correct Answer
(D) \[
3 x+3 y+7=0 \quad \& \quad 3 x+3 y-25=0
\]
Step-by-step Solution
Detailed explanation
\(\frac{1}{2} |x(2 - 5) + 1(5 - y) + (-2)(y - 2)| = 8\) \(|-3x + 5 - y - 2y + 4| = 16\) \(|-3x - 3y + 9| = 16\) \(-3x - 3y + 9 = 16 \implies 3x + 3y + 7 = 0\) \(-3x - 3y + 9 = -16 \implies 3x + 3y - 25 = 0\)
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