2 thoughts on “The mid.-points of the sides of a triangle are (-3, 2), (1, -2) and (4, 5) respectively.<br />
Find the co-ordinates of the vertic”
Step-by-step explanation:
The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time. Velocity is equivalent to a specification of an object’s speed and direction of motion
Step-by-step explanation:
The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time. Velocity is equivalent to a specification of an object’s speed and direction of motion
Answer:
Let A(x
1
y
1
),B(x
2
,y
2
)andC(x
3
,y
3
) be the vertices of △ABC.
Let D(1,2) ,E(0,-1) and F(2,-1) be the mid -points of sides BC,AC and AB
Since D is the mid point of BC.
A(x
1
y
1
),B(x
2
,y
2
)andC(x
3
,y
3
)
∴
2
x
2
+x
3
=1and
2
y
2
+y
3
=2
⟹x
2
+x
3
=2y
2
+y
3
=4→ (i)
Similarly , E and F are the mid-points of CA and Ab respectively
∴
2
x
1
+x
3
=0and
2
y
1
+y
3
=−1
⟹x
1
+x
3
=2andy
1
+y
3
−2→(2)
∴
2
x
1
+x
2
=2and
2
y
1
+y
2
=−1
⟹x
1
+x
2
=4andy
1
+y
2
=−2→(3)
From (1),(2),and(3)
(x
2
+x
3
)+(x
1
+x
3
)+(x
1
+x
2
)=2+0+4
(y
2
+y
3
)+(y
1
+y
3
)+(y
1
+y
2
)=4−2−2
⟹x
1
+x
2
+x
3
=3andy
1
+y
2
+y
3
=0→(4)
From (1) and (4)
x
1
+2=3andy
1
+4=0
⟹x
1
=1andy
1
=−4∴(1,−4)
From (2) and (4)
x
2
+0=3andy
2
−2=0
⟹(x
2
,y
2
)=B(3,2)
From (3) and (4)
x
3
+4=3andy
3
−2=0
∴(x
3
,y
3
)=C(−1,2)
∴ Coordinate of △ABC are ⇒ (1,−4);B(3,2);C(−1,2)
Step-by-step explanation:
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