AP EAMCET · Maths · Straight Lines
The centroid of the triangle formed by the lines \(x+y-1=0, x-y-1=0, x-3 y+3=0\) is
- A \(\left(\frac{4}{3}, 1\right)\)
- B \(\left(\frac{-4}{3}, 1\right)\)
- C \(\left(\frac{8}{3}, 3\right)\)
- D \(\left(\frac{-8}{3}, 3\right)\)
Answer & Solution
Correct Answer
(A) \(\left(\frac{4}{3}, 1\right)\)
Step-by-step Solution
Detailed explanation
Point of intersection of lines \(x+y-1=0\) and \(x-y-1=0\) is \(A(1,0)\). Similarly, point of intersection of lines \(x-y-1=0\) and \(x-3 y+3=0\) is \(B(3,2)\), and point of intersection of lines \(x-3 y+3=0\) and \(x+y-1=0\) is \(C(0,1)\). Now, centroid of \(\triangle A B C\)…
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