(1) Find the area, if the triangle whose vertices are
the points (3, 4), (5, 7) and (-2, -3).
Answers
Answer:
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Class 11
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>>Find the area of the triangle whose vert
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Find the area of the triangle whose vertices are (3,−4),(7,5),(−1,10)
Medium
Solution
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Given: (3,−4),(7,5),(−1,10)
Let us assume A(x
1
,y
1
)=(3,−4)
Let us assume B(x
2
,y
2
)=(7,5)
Let us assume C(x
3
,y
3
)=(−1,10)
Area of triangle
=
2
1
[x
1
(y
2
−y
3
)+x
2
(y
3
−y
1
)+x
3
(y
1
−y
2
)]
Now, Area of triangle
=
2
1
[{3(5−(10))}+(7){(10+4)}−1(−4−5)]
=
2
1
{−15+98+9}
=
2
92
=46 square units
We have,
A(x
1
,y
1
)=(2,3)
B(x
2
,y
2
)=(−2,2)
C(x
3
,y
3
)=(−1,−2)
D(x
4
,y
4
)=(3,−1)
Now,
Using distance formula we get,
AB=
(x
1
−x
2
)
2
+(y
1
−y
2
)
2
⇒AB=
(2+2)
2
+(3−2)
2
⇒AB=
17
BC=
(x
2
−x
3
)
2
+(y
2
−y
3
)
2
⇒BC=
(−2+1)
2
+(2+2)
2
⇒BC=
17
CD=
(x
3
−x
4
)
2
+(y
3
−y
4
)
2
⇒CD=
(−1−3)
2
+(−2+1)
2
⇒CD=
17
DA=
(x
4
−x
1
)
2
+(y
4
−y
1
)
2
⇒DA=
(3−2)
2
+(−1−3)
2
⇒DA=
17
Now diagonal AC=
(x
1
−x
3
)
2
+(y
1
−y
3
)
2
⇒AC=
(2+1)
2
+(3+2)
2
⇒AC=
34
And diagonal BD=
(x
2
−x
4
)
2
+(y
2
−y
3
)
4
⇒BD=
(−2−3)
2
+(2+1)
2
⇒BD=
34
So,
Sides AB=BC=CD=DA and diagonals AC=BD
Hence, the vertices are of a square ABCD.
Hence proved.