COORDINATE GEOMETRY
CLASS 10
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•Please answer all the given question in the picture.
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THANKS...
Answers
Answer:
Given:
The points are,
A(-1,-4) B(b,c) C(5,-1)
And, 2b+C=4
To find:
The value of a and b=?
Solution:
Since they are collinear,
Area of ∆ABC=0
→|x1(y2-y3)+x2(y3-y1)+x3(y1-y2)|=0
→|-1(c+1)+b(-1+4)+5(-4-c)|=0
→|-c-1+3b-20-5c|=0
→|3b-6c-21|=0
→b-2c-7=0
→b-2c=7
As we know that,
2b+c=4
Solve,
-5c=10
c=-2
Substitute c=-2 in equation 1
b-2(-2)=7
b+4=7
b=3
So,The value of b=3 and c=-2
Step-by-step explanation:
Hope it helps you.....
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.