TS EAMCET · Maths · Straight Lines
\(\mathrm{O}(0,0), \mathrm{B}(-3,-1), \mathrm{C}(-1,-3)\) are vertices of a triangle OBC. D is a point on OC and E is a point on OB. If the equation of DE is \(2 x+2 y+\sqrt{2}=0\), then the ratio in which the line DE divides the altitude of the triangle OBC is
- A \(\sqrt{2}: 4 \sqrt{2}+2\)
- B \(1: 4 \sqrt{2}+1\)
- C \(\sqrt{2}: 4 \sqrt{2}-2\)
- D \(1: 4 \sqrt{2}-1\)
Answer & Solution
Correct Answer
(D) \(1: 4 \sqrt{2}-1\)
Step-by-step Solution
Detailed explanation
Equation of line BC: \(m_{BC} = \frac{-3 - (-1)}{-1 - (-3)} = -1\). \(y + 1 = -1(x + 3) \Rightarrow x + y + 4 = 0\). Equation of altitude from O(0,0) to BC: Perpendicular to BC, so \(m_{alt} = 1\). \(y - 0 = 1(x - 0) \Rightarrow y = x\). Foot of altitude F: Intersection of…
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