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Byju's Answer
Standard XII
Mathematics
Cramer's Rule
Determine alg...
Question
Determine algebraically, the vertices of thetriangle formed dy the lines.3x-y=3,2x-3y=2, x+2y=8
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Solution
The
given
equations
are
:
3
x
-
y
=
3
.
.
.
.
.
.
.
.
.
.
.
1
2
x
-
3
y
=
2
.
.
.
.
.
.
.
.
.
.
.
2
x
+
2
y
=
8
.
.
.
.
.
.
.
.
.
.
.
.
3
Multiplying
1
by
2
and
2
by
3
,
we
get
,
6
x
-
2
y
=
6
.
.
.
.
.
.
.
.
.
.
.
.
.
4
6
x
-
9
y
=
6
.
.
.
.
.
.
.
.
.
.
.
.
.
5
Subtracting
5
from
4
,
we
get
,
7
y
=
0
⇒
y
=
0
Putting
y
=
0
in
4
,
we
get
,
6
x
-
2
×
0
=
6
⇒
6
x
-
0
=
6
⇒
6
x
=
6
⇒
x
=
1
Hence
,
the
point
of
intersection
of
1
and
2
is
1
,
0
Now
multiplying
2
by
1
and
3
by
2
,
we
get
,
2
x
-
3
y
=
2
.
.
.
.
.
.
.
.
.
.
.
6
2
x
+
4
y
=
16
.
.
.
.
.
.
.
.
.
.
.
.
.
7
Subtracting
7
from
6
,
we
get
,
-
7
y
=
-
14
⇒
y
=
2
putting
y
=
2
in
6
,
we
get
,
2
x
-
3
×
2
=
2
⇒
2
x
-
6
=
2
⇒
2
x
=
2
+
6
⇒
2
x
=
8
⇒
x
=
4
Hence
point
of
intersection
of
2
and
3
is
4
,
2
Now
,
multiplying
1
by
1
and
3
by
3
,
we
get
,
3
x
-
y
=
3
.
.
.
.
.
.
.
.
.
.
.
.
.
.
8
3
x
+
6
y
=
24
.
.
.
.
.
.
.
.
.
.
.
.
9
Subtracting
9
from
8
,
we
get
,
-
7
y
=
-
21
⇒
y
=
3
putting
y
=
3
in
8
,
we
get
,
3
x
-
3
=
3
⇒
3
x
=
3
+
3
⇒
3
x
=
6
⇒
x
=
2
Hence
,
point
of
intersection
of
1
and
3
is
2
,
3
Hence
,
vertices
of
△
are
1
,
0
,
4
,
2
and
2
,
3
Suggest Corrections
1
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Q.
Question 5
Determine, algebraically, the vertices of the triangle formed by the lines
3x – y = 3
2x – 3y = 2
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Q.
Determine algebrically the vertex of the triangle formed by lines
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