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How to find point in line $4x+3y-12=0$ whose distance from point $(-1,-2)$ and $(1,4)$ is the same. I really have no idea about that.I would like to know what to do in these kind of problems rather than giving a solution.

mathreadler
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trying to learn
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  • Hint: the locus of points equidistant from $P,Q$ is the perpendicular bisector of the segment $\overline {PQ}$. – lulu Apr 17 '17 at 20:43
  • Convert the implicit equation $4x+3y-12$ into a parametric expression $x=t; y=\tfrac{-4t}{3}+4$ ; then espress the distances by formulas $AM=\sqrt{(x_M-x_A)^2+(y_M-y_A)^2}$ and the same for $BM$. Equate them, you obtain a quadratic in $t$ that you solve. – Jean Marie Apr 17 '17 at 20:46
  • @lulu 2 You mean that you first compute the perpendicular bissector (B), then you have to find the intersection of (B) with the given line. – Jean Marie Apr 17 '17 at 20:46
  • @JeanMarie It is an easy matter to get the intersection of the perpendicular bisector and the given line. – lulu Apr 17 '17 at 20:49
  • Thank you for answering guys! I solved it . – trying to learn Apr 17 '17 at 20:59

2 Answers2

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Hint:

The points equidistant from the two points $A=(-1,-2)$ and $B=(1,4)$ are points of the perpendicular bisector of the segment $AB$. And there is one common point of this bisector with the given line.

Emilio Novati
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we want

$$( x+1 )^2+(y+2 )^2=$$ $$( x-1 )^2+( y-4 )^2$$

and $$4x+3y=12$$

the first equality gives

$$1+2x+4+4y=1-2x+16-8y $$

or $$x=3-3y $$ thus after replacing $$y=0\;\;,\;x=3$$

the common distance is $$\sqrt {20} .$$

hamam_Abdallah
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