Suppose

Suppose you know that the length of a line segment is 15, x2=6, y2=14 and x1= -3. Find the possible value of y1. Is there more than one possible answer? Why or why not?

Correct result:

y11 =  26
y12 =  2

Solution:

x2=6 y2=14 x1=3  d=15 (x1x2)2+(y1y2)2=d2 (36)2+(y114)2=152 (36)2+(y114)2=152  (36)2+(q14)2=152  q228q+52=0  a=1;b=28;c=52 D=b24ac=2824152=576 D>0  q1,2=b±D2a=28±5762 q1,2=28±242 q1,2=14±12 q1=26 q2=2   Factored form of the equation:  (q26)(q2)=0  y11=q1=26

Our quadratic equation calculator calculates it.

y12=q2=2



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Showing 1 comment:
#
Matematik
we make circle k with centre S(x2,y2) and radius r = 15 . Then we make vertical line x= -3 . It make two intersections with circle k thus solutions are two: y11,y12.

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