AP EAMCET · Maths · Straight Lines
Identify the point on the line 2x + 3y + 7 = 0, which is at a distance of + 3 units from (1, − 3).
- A \(\left(\frac{\sqrt{13}+9}{\sqrt{13}}, \frac{-3 \sqrt{13}+6}{\sqrt{13}}\right)\)
- B \(\left(\frac{\sqrt{13}-9}{\sqrt{13}}, \frac{-3 \sqrt{13}-6}{\sqrt{13}}\right)\)
- C \(\left(\frac{\sqrt{13}-9}{\sqrt{13}}, \frac{-3 \sqrt{13}+6}{\sqrt{13}}\right)\)
- D \(\left(\frac{\sqrt{13}+9}{\sqrt{13}}, \frac{3 \sqrt{13}-6}{\sqrt{13}}\right)\)
Answer & Solution
Correct Answer
(C) \(\left(\frac{\sqrt{13}-9}{\sqrt{13}}, \frac{-3 \sqrt{13}+6}{\sqrt{13}}\right)\)
Step-by-step Solution
Detailed explanation
Let \(P(\alpha, \beta)\) be the point on the line 2x + 3y + 7 = 0 \(\begin{array}{rlrl} & \Rightarrow & 2 \alpha+3 \beta+7 & =0 \\ & \beta & =\left(\frac{-7-2 \alpha}{3}\right) \\ \therefore & & P & =\left(\alpha, \frac{-7-2 \alpha}{3}\right)\end{array}\) Given, point A = (1, −…
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