Question

the mid points of the sides of a triangle are (2,1) (-5,7) and (-5,-5). find the equations of the sides of the triangle.​

Answer 1

Answer:

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Answer 2

The equations of the sides of $\triangle ABC$ is $x_{1} +x_{2}=4$,$y_{1} +y_{2}=2$,$x_{2} +x_{3}=-10$,$y_{2} +y_{3}=14$,$x_{1} +x_{3}=-10$ and $y_{1} +y_{3}=-10$

Step-by-step explanation:

Given,

Mid points of the sides of a triangle are $(2,1),(-5,7)$ and $(-5,-5)$

Let,

For $\triangle ABC$ ,

Point $A$ is $(x_{1} ,y_{1} )$ , Point $B$ is $(x_{2},y_{2})$  and Point $C$ is $(x_{3},y_{3} )$

∴ Mid point of sides $AB$ is $(\frac{x_{1} +x_{2} }{2},\frac{y_{1}+y_{2} }{2} )$

From given data this point is similar with $(2,1)$

∴ $\frac{x_{1} +x_{2} }{2}=2$ and $\frac{y_{1}+y_{2} }{2} =1$

⇒ $x_{1} +x_{2}=4$ and $y_{1} +y_{2}=2$

And Mid point of sides $BC$ is $(\frac{x_{3} +x_{2} }{2},\frac{y_{3}+y_{2} }{2} )$

From given data this point is similar with $(-5,7)$

Similarly,

$x_{2} +x_{3}=-10$ and $y_{2} +y_{3}=14$

Also Mid point of sides $AC$ is $(\frac{x_{3} +x_{1} }{2},\frac{y_{3}+y_{1} }{2} )$

From given data this point is similar with $(-5,-5)$

Similarly,

$x_{1} +x_{3}=-10$ and $y_{1} +y_{3}=-10$

∴ The equations of the sides of $\triangle ABC$ is $x_{1} +x_{2}=4$,$y_{1} +y_{2}=2$,$x_{2} +x_{3}=-10$,$y_{2} +y_{3}=14$,$x_{1} +x_{3}=-10$ and $y_{1} +y_{3}=-10$