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 =  0

Solution:

$$\begin{gathered} d=\mathrm{\mid }AB\mathrm{\mid } \\ d=\sqrt{(-3-9{)}^2+(-5-(-10){)}^2}=13 \\ {d}^2=(-3-2{)}^2+(-5-k{)}^2 \\ 13^2=(-3-2{)}^2+(-5-k{)}^2 \\ -k^2-10k+119=0 \\ {k}^2+10k-119=0 \\ a=1;b=10;c=-119 \\ D=b^2-4ac=10^2-4\cdot 1\cdot (-119)=576 \\ D>0 \\ {k}_{1,2}={\displaystyle \frac{-b\pm \sqrt{D}}{2a}}={\displaystyle \frac{-10\pm \sqrt{576}}{2}} \\ {k}_{1,2}={\displaystyle \frac{-10\pm 24}{2}} \\ {k}_{1,2}=-5\pm 12 \\ {k}_1=7 \\ {k}_2=-17 \\ \text{ Factored form of the equation: } \\ (k-7)(k+17)=0 \\ {k}_1=7 \end{gathered}$$

Our quadratic equation calculator calculates it.

$k_2=0$

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