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KEAM 2017 Math Paper
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© examsnet.com
Question : 38
Total: 120
The distance of the point (3, – 5) from the line 3x – 4y – 26 = 0 is
3/7
2/5
7/5
3/5
1
Validate
Solution:
Distance of a point
(
x
1
,
y
1
)
from the line
A
x
+
B
y
+
C
=
0
is given by
=
|
A
x
1
+
B
y
1
+
C
√
A
2
+
B
2
|
∴
Distance of the point (3,-5) from the line
3
x
−
4
y
−
26
=
0
is equal to
=
∣
3
×
3
−
4
×
(
−
5
)
−
26
√
(
3
)
2
+
(
−
4
)
2
=
|
9
+
20
−
26
√
9
+
16
|
=
|
3
5
|
=
3
5
unit
© examsnet.com
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