document.write( "Question 198664: the vertices of the base of an isosceles triangle are (1,2) & (4,-1). find the ordinate of the third vertex if its abcissa is 6 \n" ); document.write( "

Algebra.Com's Answer #149212 by Alan3354(69443) $\"\"$ $\"About$

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the vertices of the base of an isosceles triangle are (1,2) & (4,-1). find the ordinate of the third vertex if its abcissa is 6
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\n" ); document.write( "The 3rd vertex has to be equidistant from the 2 given points, and it's on the line x = 6. The point will be (6,y).
\n" ); document.write( "The distance from (1,2) is sqrt of ((6-1)^2 + (y-2)^2)
\n" ); document.write( "The distance from (4,-1) is sqrt of ((6-4)^2 + (y+1)^2)
\n" ); document.write( "The distances are equal, so the squares are equal.
\n" ); document.write( "5^2 + (y-2)^2 = 2^2 + (y+1)^2
\n" ); document.write( "y^2 - 4y + 29 = y^2 + 2y + 5
\n" ); document.write( "-6y = -24
\n" ); document.write( "y = 4
\n" ); document.write( "Vertex at (6,4)
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