Three points

Three points A (-3;-5) B (9;-10) and C (2;k) . AB=AC
What is value of k?

Correct result:

k1 =  7
k2 =  -17

Solution:

d=∣AB∣ d=(−3−9)2+(−5−(−10))2=13 d2=(−3−2)2+(−5−k)2  132=(−3−2)2+(−5−k)2  −k2−10k+119=0 k2+10k−119=0  a=1;b=10;c=−119 D=b2−4ac=102−4⋅1⋅(−119)=576 D>0  k1,2=−b±D2a=−10±5762 k1,2=−10±242 k1,2=−5±12 k1=7 k2=−17   Factored form of the equation:  (k−7)(k+17)=0  k1=7d=|AB| \ \\ d=\sqrt{ (-3-9)^2+(-5-(-10))^2 }=13 \ \\ d^2=(-3-2)^2+(-5-k)^2 \ \\ \ \\ 13^2=(-3-2)^2+(-5-k)^2 \ \\ \ \\ -k^2 -10k +119=0 \ \\ k^2 +10k -119=0 \ \\ \ \\ a=1; b=10; c=-119 \ \\ D=b^2 - 4ac=10^2 - 4\cdot 1 \cdot (-119)=576 \ \\ D>0 \ \\ \ \\ k_{1,2}=\dfrac{ -b \pm \sqrt{ D } }{ 2a }=\dfrac{ -10 \pm \sqrt{ 576 } }{ 2 } \ \\ k_{1,2}=\dfrac{ -10 \pm 24 }{ 2 } \ \\ k_{1,2}=-5 \pm 12 \ \\ k_{1}=7 \ \\ k_{2}=-17 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (k -7) (k +17)=0 \ \\ \ \\ k_{1}=7

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k2=(−17)=−17

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