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Points A, B and C are the three vertices of an isosceles triangle in which AC=BC. The coordinates of A and B are (-2,1) and (6,-3) respectively and the equation of AC is 4x-7y+15=0.
\n" ); document.write( "Let the coordinates of C be (h,k).
\n" ); document.write( "We are given that AC=BC
\n" ); document.write( "Squaring both sides we get AC^2=BC^2
\n" ); document.write( "The distance between two points(x1,y1) and (x2,y2) is given by sqrt((y2-y1)^2+(x2-x1)^2)
\n" ); document.write( "Plugging in the values we get, AC^2=(h+2)^2+(k-1)^2=h^2+4h+4+k^2-2k+1=h^2+k^2+4h-2k+5
\n" ); document.write( "We also have BC^2=(h-6)^2+(k+3)^2=h^2-12h+36+k^2+6k+9=h^2+k^2-12h+6k+45
\n" ); document.write( "Equating both we get,
\n" ); document.write( "h^2+k^2+4h-2k+5=h^2+k^2-12h+6k+45
\n" ); document.write( "4h-2k+5=-12h+6k+45
\n" ); document.write( "16h-8k=40
\n" ); document.write( "Dividing by 8 on both sides we get,
\n" ); document.write( "2h-k=5................................Equation 1
\n" ); document.write( "Since (h,k) lies on the line AC we get,
\n" ); document.write( "4h-7k+15=0----------------------------Equation 2
\n" ); document.write( "multiplying Equation 1 by 7 we get,
\n" ); document.write( "14h-7k=35.............................Equation 3\r
\n" ); document.write( "\n" ); document.write( "Subtracting equation 2 from 3 we get,
\n" ); document.write( "14h-7k-4h+7k-15=35
\n" ); document.write( "Simplifying we get,
\n" ); document.write( "10h=50
\n" ); document.write( "h=5
\n" ); document.write( "Substituting in equation 1 we get,
\n" ); document.write( "10-k=5
\n" ); document.write( "k=5
\n" ); document.write( "The point C has coordinates (5,5)
\n" ); document.write( "AC^2=49+16=65 square units
\n" ); document.write( "BC^2=1=64=65 square units
\n" ); document.write( "The point (5,5) also satisfies the equation 4x-7y+15=0
\n" ); document.write( "Hence C(5,5) is the correct solution. \n" ); document.write( "