# 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:

$d=|AB| \ \\ d=\sqrt{ (-3-9)^2+(-5-(-10))^2 }=13 \ \\ d^2=(-3-2)^2+(-5-k)^2 \ \\ \ \\ 13^2=(-3-2)^2+(-5-k)^2 \ \\ \ \\ -k^2 -10k +119=0 \ \\ k^2 +10k -119=0 \ \\ \ \\ a=1; b=10; c=-119 \ \\ D=b^2 - 4ac=10^2 - 4\cdot 1 \cdot (-119)=576 \ \\ D>0 \ \\ \ \\ k_{1,2}=\dfrac{ -b \pm \sqrt{ D } }{ 2a }=\dfrac{ -10 \pm \sqrt{ 576 } }{ 2 } \ \\ k_{1,2}=\dfrac{ -10 \pm 24 }{ 2 } \ \\ k_{1,2}=-5 \pm 12 \ \\ k_{1}=7 \ \\ k_{2}=-17 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (k -7) (k +17)=0 \ \\ \ \\ k_{1}=7$

Checkout calculation with our calculator of quadratic equations.

${k}_{2}=\left(-17\right)=-17$

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