Geo2.32

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

Let the point be (a,b)

Given it is equidistant from the two points,so

{\sqrt  {(a+5)^{2}+(b-6)^{2}}}={\sqrt  {(a-3)^{2}+(b-2)^{2}}}\,

a^{2}+25+10a+b^{2}+36-12b=a^{2}+9-6a+b^{2}+4-4b\,

Simplifying we get,

16a-8b=-48\,

2a-b=-6\,

Since the point lies on the given equation,

3a+b=-4\,

Solving these two equations,

a=-2,b=2\,

Hence the point is (-2,2)


Main Page:Geometry:Straight Lines-I