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Solution 4.1:4c

From Förberedande kurs i matematik 1

Revision as of 10:15, 27 September 2008 by Ian (Talk | contribs)
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Let the point on the x-axis have coordinates x0 , where x is an unknown number. Then, using the distance formula, the distance from the point x0  to 33  and 51  is given by


x32+032  and x52+012 


Because these two distances should be the same, we get the equation


x32+9=x52+1 


or, if we take the square, so as to get rid of the square root sign,


x32+9=x52+1 


Now, expand the squares and collect together all the terms onto one side:


x26x+9+9=x210x+25+14x8=0


This gives that x=2, i.e. the point on the x-axis is 20 .


As a final step, we check that we have calculated correctly and that the distances really are the same. The distance between 20  and 33  is


322+302=12+32=1+9=10 


and the distance between 20  and 51  is


522+102=32+12=9+1=10 


NOTE: Although we squared our root equation, it is not in fact necessary to test the solution for that reason, because the expressions under the root signs are sums of squares which are never negative and therefore cannot give rise to so-called false roots.