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Byju's Answer
Standard X
Mathematics
Collinearity Condition
The area of Δ...
Question
The area of ∆ABC with vertices A(1, −1), B(−4, 6) and C(−3, −5) is
(a) 52 sq units
(b) 27 sq units
(c) 48 sq units
(d) 24 sq units
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Solution
(d) 24 sq units
The given points are
A
1
,
-
1
,
B
-
4
,
6
and
C
-
3
,
-
5
.
Here,
x
1
=
1
,
y
1
=
-
1
,
x
2
=
-
4
,
y
2
=
6
and
x
3
=
-
3
,
y
3
=
-
5
Therefore,
Area of
∆
A
B
C
=
1
2
x
1
y
2
-
y
3
+
x
2
y
3
-
y
1
+
x
3
y
1
-
y
2
=
1
2
1
6
+
5
+
-
4
-
5
+
1
+
-
3
-
1
-
6
=
1
2
11
+
16
+
21
=
1
2
×
48
=
24
sq units
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Q.
The area of ∆AOB having vertices A(0, 6), O(0, 0) and B(6, 0) is
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(b) 36 sq units
(c) 18 sq units
(d) 24 sq units