TS EAMCET · Maths · Straight Lines
If \(\mathrm{P}\) is a point equidistant from all the vertices \(\mathrm{A}(-1,3)\), \(\mathrm{B}(3,5), \mathrm{C}(5,7)\) of a triangle \(\mathrm{ABC}\) then \(\mathrm{PA}=\)
- A 11
- B \(\sqrt{140}\)
- C 13
- D \(\sqrt{130}\)
Answer & Solution
Correct Answer
(D) \(\sqrt{130}\)
Step-by-step Solution
Detailed explanation
Given vertices of the triangle \({ABC} A(-1,3), {B}(3,5)\) and \({C}(5,7)\) According to the question, \({PA}={PB}={PC}\) \( \begin{aligned} & \sqrt{(x+1)^2+(y-3)^2}=\sqrt{(x-3)^2+(y-5)^2}= \\ & \sqrt{(x-5)^2+(y-7)^2} \end{aligned} \) Take square both sides,…
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