TS EAMCET · Maths · Straight Lines
The point on the line \(3 x+4 y=5\) which is equidistant from \((1,2)\) and \((3,4)\) is
- A \((7,-4)\)
- B \((15,-10)\)
- C \((1 / 7,8 / 7)\)
- D \((0,5 / 4)\)
Answer & Solution
Correct Answer
(B) \((15,-10)\)
Step-by-step Solution
Detailed explanation
Let point \(\left(x_1, y_1\right)\) be on the line \(3 x+4 y=5\). \(\therefore \quad 3 x_1+4 y_1=5\) ...(i) Also, \(\quad\left(x_1-1\right)^2+\left(y_1-2\right)^2=\left(x_1-3\right)^2\)…
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