Three points

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

Result

k1 =  7
k2 =  -17

Solution:

d=∣AB∣ d=(−3−9)2+(−5−(−10))2=13 d2=(−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=k1=7d=|AB| \ \\ d=\sqrt{ (-3-9)^2+(-5-(-10))^2 }=13 \ \\ d^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}=k_{ 1 }=7

Checkout calculation with our calculator of quadratic equations.

k2=k2=(−17)=−17k_{2}=k_{ 2 }=(-17)=-17

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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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