Isosceles triangle

In an isosceles triangle ABC with base AB; A [3,4]; B [1,6] and the vertex C lies on the line 5x - 6y - 16 = 0. Calculate the coordinates of vertex C.

Correct answer:

x =  2
y =  -1

Step-by-step explanation:

$$\begin{gathered} x_0=-3;y_0=4 \\ {x}_1=1;y_1=6 \\ (x-x_0{)}^2+(y-y_0{)}^2=(x-x_1{)}^2+(y-y_1{)}^2 \\ 5\cdot x-6\cdot y-16=0 \\ y=(5\cdot x-16)\mathrm{/}6 \\ (x-x_0{)}^2+((5\cdot x-16)\mathrm{/}6-y_0{)}^2=(x-x_1{)}^2+((5\cdot x-16)\mathrm{/}6-y_1{)}^2 \\ (x+3{)}^2+((5\cdot x-16)\mathrm{/}6-4{)}^2=(x-1{)}^2+((5\cdot x-16)\mathrm{/}6-6{)}^2 \\ {x}^2+6x+9+25\mathrm{/}36x^2-100x\mathrm{/}9+400\mathrm{/}9=x^2-2x+1+25\mathrm{/}36x^2-130x\mathrm{/}9+676\mathrm{/}9 \\ 6\cdot x+9-100\cdot x\mathrm{/}9+400\mathrm{/}9=-2\cdot x+1-130\cdot x\mathrm{/}9+676\mathrm{/}9 \\ 102x=204 \\ x=2=2 \end{gathered}$$

$y=(5\cdot x-16)\mathrm{/}6=(5\cdot 2-16)\mathrm{/}6=-1$

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