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...
Answers
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
∴ 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:
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.