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AP EAMCET Engineering 2018 Apr 22 Shift 2 Paper
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© examsnet.com
Question : 45
Total: 160
If the straight line
L
≡
3
x
+
4
y
−
k
=
0
cuts the line segment joining the points P(2, -1) and Q(1,1) in the ration 4:1, then the equation of the parallel to the line
y
=
x
and concurrent with the lines PQ and L = 0 is
2
x
−
2
y
+
7
=
0
x
−
y
+
1
=
0
5
x
−
5
y
−
3
=
0
y
=
x
+
3
Validate
Solution:
The point of intersection of line L = 0 and line segment PQ is,
R
(
(
4
×
1
)
+
(
1
×
2
)
5
,
(
4
×
1
)
+
(
1
×
(
−
1
)
)
5
)
=
R
(
6
5
,
3
5
)
For point R on line L = 0 is,
18
5
+
12
5
−
k
=
0
k
=
6
Now, equation of line concurrent with the lines L = 0 and PQ,
(
3
x
+
4
y
−
6
)
+
λ
(
2
x
+
y
−
3
)
=
0
The above line is parallel to x = y , so
−
3
+
2
λ
4
+
λ
=
1
λ
=
−
7
3
So, the equation of the line is,
5
x
−
5
y
−
3
=
0
© examsnet.com
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