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Q.Find the point on the straight line 3x+y+4=03x + y + 4 = 0 which is equidistant from the points (−5,6)(-5, 6) and (3,2)(3, 2).

Andhra Pradesh BieapBIEAP Intermediate Board (1st Year) 2023Subjective· 4mImportance★★★★★
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A point equidistant from two points lies on their perpendicular bisector; intersect that bisector with the given line.

Let the required point be equidistant from A(−5,6)A(-5,6) and B(3,2)B(3,2), so it lies on the perpendicular bisector of ABAB.

Midpoint of AB=(−5+32,6+22)=(−1,4)AB = \left(\dfrac{-5+3}{2},\dfrac{6+2}{2}\right) = (-1,4).

Slope of AB=2−63−(−5)=−48=−12AB = \dfrac{2-6}{3-(-5)} = \dfrac{-4}{8} = -\dfrac12, so the perpendicular bisector has slope 22.

Perpendicular bisector: y−4=2(x+1)⇒y=2x+6y-4 = 2(x+1) \Rightarrow y = 2x+6.

Solve simultaneously with the given line 3x+y+4=0⇒y=−3x−43x+y+4=0 \Rightarrow y=-3x-4:

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