COMEDK · Maths · 10. Straight Lines
Let \(\mathrm{ABC}\) be a triangle with equations of its sides \(\mathrm{AB}, \mathrm{BC}\). \(\mathrm{CA}\) respectively are \(x-2=0, y-5=0\) and \(5 x+2 y-10=0\). Then the orthocentre of triangle lies on the line
- A \(x-2 y=1\)
- B \(x-y=0\)
- C \(3 x+y=1\)
- D \(4 x+y=13\)
Answer & Solution
Correct Answer
(D) \(4 x+y=13\)
Step-by-step Solution
Detailed explanation
The equations of the sides are \(L_1: x - 2 = 0\), \(L_2: y - 5 = 0\), and \(L_3: 5x + 2y - 10 = 0\). Find the vertices by solving the equations pairwise: Vertex A is the intersection of \(L_1\) and \(L_3\): \(x = 2\), \(5(2) + 2y - 10 = 0 \Rightarrow 2y = 0 \Rightarrow y = 0\).…
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