Question

1.The midpoints of the sides of a triangle are (5,1), (3,-5) and (-5, -1).
Find the coordinates of the Vertices of the triangle.

please give me the answer. ​

Answer 1

Answer:

The coordinates of the sides of the triangle are A (-3, 5), B (13, -3), and C (5, -7)

Step-by-step explanation:

Given: The midpoints of the sides of a triangle are (5,1), (3,-5), and (-5, -1).

Let the midpoint of AB, BC, and AC be (5,1), (3,-5), and (-5, -1) respectively.

To Calculate: The coordinates of the Vertices of the triangle.

Solution:

Let the coordinates of the vertices of the triangles A, B and C be

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

So, Using the midpoint formula, we get

$\frac{x_{1} +x_{2} }{2} = 5 \\x_{1} +x_{2} = 10$   equation (1)

And

$\frac{y_{1} +y_{2} }{2} =1 \\y_{1} +y_{2} = 2$  equation (2)

Similarly, we have :

$x_{2} +x_{3} = 6$    equation (3)

$y_{2} +y_{3} =-10$   equation (4)

And,

$x_{1} +x_{3} = -10$  equation (5)

$\\\\y_{1} +y_{3} = -2$  equation (6)

Now,

Adding equations 1, 3 and 5

And Adding equations 2, 4 and 6

We get,

$2(x_{1} +x_{3} + x_{3} )= 6$

$(x_{1} +x_{3} + x_{3} )= 3$   equation (7)

And,

$2(y_{1} +y_{2} + y_{3} )= -10$

$(x_{1} +x_{3} + x_{3} )= -5$   equation (8)

Now, On Solving these equations, we get

On, (7) - (1)

$x_{3} = -7$

On, (7) - (3)

$x_{1} = -3$

On, (7) - (5)

$x_{2} = 13$

Now,

On, (8) - (2)

$y_{3} = -7$

On, (8) - (4)

$y_{1} = 5$

On, (8) - (6)

$y_{2} = -3$

Thus, the coordinates of the sides of the triangle are A (-3, 5), B (13, -3), and C (5, -7)

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Answer 2

Answer:

The vertex of ABC triangle is A=(7,6), B=(23,-2), C=(3,-6).

Step-by-step explanation:

As per question we need to find the vertex of the triangle.

Let the triangle is ABC. A,B,C is the vertex of the triangle.

The given data is midpoints of the sides of a triangle are (5,1), (3,-5) and (-5, -1).

Let the point (5,1) is the middle point of the side AB, and the point (3,-5) is the middle point of the side BC, and the point (-5,-1).

Now also let the coordinates of the vertex is A=(x₁,y₁), B=(x₂,y₂), C=(x₃,y₃).

we know that the middle point of the side AB  is $\frac{(x1+x2)}{2} =5, \frac{(y1+y2)}{2} =1$

and the middle point of BC is $\frac{x2+x3}{2}=3, \frac{(y2+y3)}{2}=-5$ and the middle point of AC is $\frac{(x3+x1)}{2} =-5, \frac{y3+y1}{2} = -1$.

now from AB, BC, CA sides we get

x₁ +x₂ =10.......... (1),

y₁+y₂ =2............ (2),

x₂ +x₃ =6.........(3),

y₂ +y₃ = -10........(4),

x₃+x₁=-10..........(5),

y₃ +y₁= -2...........(6)

add equation (1), (3), and (5) we get

2(x₁ +x₂ + x₃) =26

or, (x₁ +x₂ + x₃)=13......(7)

now put the x₁ +x₂ =10 in equation (7) we get

x₃=3,

now put the x₂ +x₃ =6 in equation (7) we get

x₁=7

now put the x₃+x₁=-10 in equation (7) we get

x₂=23

continuing this process we get

y₁+y₂+y₃=-4......(8)

from (8) we get

y₁=6

y₂=-2

y₃ =-6

Therefore the vertex of ABC triangle is A=(7,6), B=(23,-2), C=(3,-6).

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