# 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?

Result

y11 =  26
y12 =  2

#### Solution:

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

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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. Tips to related online calculators
For Basic calculations in analytic geometry is helpful line slope calculator. From coordinates of two points in the plane it calculate slope, normal and parametric line equation(s), slope, directional angle, direction vector, the length of segment, intersections the coordinate axes etc.
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