Question

In a triangle ABC (4,2) is the midpoint of AB.The midpoint of BC is (5,4). midpoint of AC (3,3).Find vertices of triangle.​

Answer 1

✏ Step by step explanation

✬Let us assume a triangle ABC such that D, E and F are the midpoints of AB, BC and CA respectively.

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✬Since, it is given that (4,2) is the midpoint of AB.

✬This implies, coordinates of D are (4, 2).

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✬The midpoint of BC is (5,4).

✬This implies coordinates of E are (5, 4)

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✬Also, The midpoint of AC is (3,3)

✬This implies, coordinates of F are (3, 3).

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✬ Let assume that coordinates of vertices are A(a, b), B(c, d) and C(e, f).

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$\begin{gathered}\begin{gathered}\bf To \: Find :- \begin{cases} &\sf{Vertices \: of \: A, \: B, \: and \: C.} \end{cases}\end{gathered}\end{gathered}$

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$\large\underline{\bold{Solution :- }}$

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$\large\underline{\bold{❥︎Step :- 1 }}$

 ✬ Since, D(4, 2) is the midpoint of line segment joinjng A(a, b) and B(c, d).

So, by using midpoint Formula, we get

$\sf \: \sf \:  ⟼ (\dfrac{a + c}{2} , \dfrac{b + d}{2} ) = (4 , 2)$

 ✬ On comparing, we get

$\bf\implies \:\dfrac{a + c}{2} = 4 \: and \: \dfrac{b + d}{2} = 2$

$\bf\implies \:a + c = 8 - - (1) \: and \: b + d = 4 - - (2)$

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$\large\underline{\bold{❥︎Step :- 2 }}$

✬ Since, E(5, 4) is the midpoint of line segment joinjng C(e, f) and B(c, d).

So, by using midpoint Formula, we get

$\sf \:  ⟼\sf \:  (\dfrac{c + e}{2} , \dfrac{d + f}{2} ) = (5 ,4 )$

 ✬ On comparing, we get

$\bf\implies \:\dfrac{c + e}{2} = 5 \: and \: \dfrac{d + f}{2} = 4$

$\bf\implies \:c + e = 10 \: - - (3)and \: d + f = 8 - - (4)$

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$\large\underline{\bold{❥︎Step :- 3 }}$

✬ Since, E(3, 3) is the midpoint of line segment joinjng A(a, b) and C(e, f).

So, by using midpoint Formula, we get

$\sf \:  ⟼\sf \:  (\dfrac{a + e}{2} , \dfrac{b + f}{2} ) = (3 , 3)$

 ✬ On comparing, we get

$\bf\implies \:\dfrac{a + e}{2} = 3 \: and \: \dfrac{b + f}{2} = 3$

$\bf\implies \:a + e = 6 - - (5) \: and \: b + f = 6 - - (6)$

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$\large\underline{\bold{❥︎Step :- 4 }}$

 ✬ Adding equations (1), (3) and (5), we get

$\bf \:  ⟼ 2( a+ c+ e) = 24$

$\bf\implies \: \: a +c + e = 12 - - - (7)$

 ✬On Subtracting equation (1) from (7), we get

$\bf \:  ⟼ e \: = \: 4 - - - (8)$

✬On Subtracting equation (2) from (7), we get

$\bf \:  ⟼ a \: = \: 2 - - - (9)$

✬On Subtracting equation (3) from (7), we get

$\bf \:  ⟼ c \: = \: 6 - - - (10)$

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$\large\underline{\bold{❥︎Step :- 5 }}$

✬ Adding equations (2), (4) and (6), we get

$\bf\implies \:2(b +d + f) = 18$

$\bf \:  ⟼ \: b+ d + f \: = 9 - - - (11)$

✬On Subtracting equation (2) from (11), we get

$\bf \:  ⟼ f \: = 5 - - - (12)$

✬On Subtracting equation (4) from (11), we get

$\bf \:  ⟼ b \: = 1 - - - (13)$

✬On Subtracting equation (6) from (11), we get

$\bf \:  ⟼ d \: = 3 - - - (14)$

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$\begin{gathered}\begin{gathered}\bf So, \: vertices \: are - \begin{cases} &\sf{A(2,1)} \\ &\sf{B(6,3)}\\ &\sf{C(4,5)} \end{cases}\end{gathered}\end{gathered}$

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