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=26x_{2}=6 \ \\ y_{2}=14 \ \\ x_{1}=-3 \ \\ \ \\ d=15 \ \\ (x_{1}-x_{2})^2 +(y_{1}-y_{2})^2=d^2 \ \\ (-3-6)^2 +(y_{1}-14)^2=15^2 \ \\ (-3-6)^2 +(y_{1}-14)^2=15^2 \ \\ \ \\ (-3-6)^2 +(q-14)^2=15^2 \ \\ \ \\ q^2 -28q +52=0 \ \\ \ \\ a=1; b=-28; c=52 \ \\ D=b^2 - 4ac=28^2 - 4\cdot 1 \cdot 52=576 \ \\ D>0 \ \\ \ \\ q_{1,2}=\dfrac{ -b \pm \sqrt{ D } }{ 2a }=\dfrac{ 28 \pm \sqrt{ 576 } }{ 2 } \ \\ q_{1,2}=\dfrac{ 28 \pm 24 }{ 2 } \ \\ q_{1,2}=14 \pm 12 \ \\ q_{1}=26 \ \\ q_{2}=2 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (q -26) (q -2)=0 \ \\ \ \\ y_{11}=q_{1}=26

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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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