Question

COORDINATE GEOMETRY
CLASS 10
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•Please answer given question in the picture.
i.e.question no.6
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THANKS...​ ​

Answer 1

We know that the median of a triangle is drawn from a vertex to the midpoint of the side opposite to the vertex, so first, let's find the midpoint of BC.

Finding the Midpoint of BC:

B(3, -2) & C(5, 2)

x₁ = 3

x₂ = 5

y₁ = -2

y₂ = 2

$\sf \Longrightarrow D(x, y) = \Bigg( \dfrac{x_1 + x_2}{2} , \dfrac{y_1 + y_2}{2} \Bigg)$

$\sf \Longrightarrow D(x, y) = \Bigg( \dfrac{3 + 5}{2} , \dfrac{-2 + 2}{2} \Bigg)$

$\sf \Longrightarrow D(x, y) = \Bigg( \dfrac{8}{2} , \dfrac{0}{2} \Bigg)$

$\sf \Longrightarrow D(x, y) = \Big(4 , 0\Big)$

∴ Therefore the midpoint of BC is (4, 0).

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Now, let's find out the area of ACD & ABD and see if their areas are equal.

Area of ACD.

A(4, -6)

C(5, 2)

D(4, 0)

Area of a triangle can be written as:

ar(ΔACD) = ¹/₂ [x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂)]

ar(ΔACD) = ¹/₂ [4(2 - 0) + 5(0 - (-6)) + 4(-6 - 2)]

ar(ΔACD) = ¹/₂ [4(2) + 5(0 + 6) + 4(-8)]

ar(ΔACD) = ¹/₂ [8 + 5(6) + (-32)]

ar(ΔACD) = ¹/₂ [8 + 30 - 32]

ar(ΔACD) = ¹/₂ [8 - 2]

ar(ΔACD) = ¹/₂ [6]

ar(ΔACD) = 3 sq.units. → Statement(1)

Area of ABD.

A(4, -6)

B(3, -2)

D(4, 0)

Area of a triangle can be written as:

ar(ΔABD) = ¹/₂ [x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂)]

ar(ΔABD) = ¹/₂ [4(-2 - 0) + 3(0 - (-6)) + 4(-6 - (-2))]

ar(ΔABD) = ¹/₂ [4(-2) + 3(0 + 6) + 4(-6+ 2)]

ar(ΔABD) = ¹/₂ [-8 + 3(6) + 4(-4)]

ar(ΔABD) = ¹/₂ [-8 + 18 + -16]

ar(ΔABD) = ¹/₂ [10 + -16]

ar(ΔABD) = ¹/₂ [-6]

Area can't be negative, therefore we'll take it to be +ve.

ar(ΔABD) = 3 sq.units. → Statement(2)

From statement 1 & 2,

ar(ΔACD) = ar(ΔABD)

Hence proved.

Answer 2

Answer:

We know that the median of a triangle is drawn from a vertex to the midpoint of the side opposite to the vertex, so first, let's find the midpoint of BC.

Finding the Midpoint of BC:

B(3, -2) & C(5, 2)

x₁ = 3

x₂ = 5

y₁ = -2

y₂ = 2

\sf \Longrightarrow D(x, y) = \Bigg( \dfrac{x_1 + x_2}{2} , \dfrac{y_1 + y_2}{2} \Bigg)⟹D(x,y)=(

2

x

1

+x

2

,

2

y

1

+y

2

)

\sf \Longrightarrow D(x, y) = \Bigg( \dfrac{3 + 5}{2} , \dfrac{-2 + 2}{2} \Bigg)⟹D(x,y)=(

2

3+5

,

2

−2+2

)

\sf \Longrightarrow D(x, y) = \Bigg( \dfrac{8}{2} , \dfrac{0}{2} \Bigg)⟹D(x,y)=(

2

8

,

2

0

)

\sf \Longrightarrow D(x, y) = \Big(4 , 0\Big)⟹D(x,y)=(4,0)

∴ Therefore the midpoint of BC is (4, 0).

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Now, let's find out the area of ACD & ABD and see if their areas are equal.

Area of ACD.

A(4, -6)

C(5, 2)

D(4, 0)

Area of a triangle can be written as:

ar(ΔACD) = ¹/₂ [x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂)]

ar(ΔACD) = ¹/₂ [4(2 - 0) + 5(0 - (-6)) + 4(-6 - 2)]

ar(ΔACD) = ¹/₂ [4(2) + 5(0 + 6) + 4(-8)]

ar(ΔACD) = ¹/₂ [8 + 5(6) + (-32)]

ar(ΔACD) = ¹/₂ [8 + 30 - 32]

ar(ΔACD) = ¹/₂ [8 - 2]

ar(ΔACD) = ¹/₂ [6]

ar(ΔACD) = 3 sq.units. → Statement(1)

Area of ABD.

A(4, -6)

B(3, -2)

D(4, 0)

Area of a triangle can be written as:

ar(ΔABD) = ¹/₂ [x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂)]

ar(ΔABD) = ¹/₂ [4(-2 - 0) + 3(0 - (-6)) + 4(-6 - (-2))]

ar(ΔABD) = ¹/₂ [4(-2) + 3(0 + 6) + 4(-6+ 2)]

ar(ΔABD) = ¹/₂ [-8 + 3(6) + 4(-4)]

ar(ΔABD) = ¹/₂ [-8 + 18 + -16]

ar(ΔABD) = ¹/₂ [10 + -16]

ar(ΔABD) = ¹/₂ [-6]

Area can't be negative, therefore we'll take it to be +ve.

ar(ΔABD) = 3 sq.units. → Statement(2)

From statement 1 & 2,

ar(ΔACD) = ar(ΔABD)

Hence proved.