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Byju's Answer
Standard XII
Mathematics
Director Circle
Show that the...
Question
Show that the origin is equidistant from the lines 4x + 3y + 10 = 0; 5x − 12y + 26 = 0 and 7x + 24y = 50.
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Solution
Let us write down the normal forms of the given lines.
First line:
4x + 3y + 10 = 0
⇒
-
4
x
-
3
y
=
10
⇒
-
4
-
4
2
+
-
3
2
x
-
3
-
4
2
+
-
3
2
y
=
10
-
4
2
+
-
3
2
Dividing
both
sides
by
coefficient
of
x
2
+
coefficient
of
y
2
⇒
-
4
5
x
-
3
5
y
=
2
∴
p
=
2
Second line:
5x − 12y + 26 = 0
⇒
-
5
x
+
12
y
=
26
⇒
-
5
-
5
2
+
12
2
x
+
12
-
5
2
+
12
2
y
=
26
-
5
2
+
12
2
Dividing
both
sides
by
coefficient
of
x
2
+
coefficient
of
y
2
⇒
-
5
13
x
+
12
13
y
=
2
∴
p
=
2
Third line:
7x + 24y = 50
⇒
7
7
2
+
24
2
x
+
24
7
2
+
24
2
y
=
50
7
2
+
24
2
Dividing
both
sides
by
coefficient
of
x
2
+
coefficient
of
y
2
⇒
7
25
x
+
24
25
y
=
2
∴
p
=
2
Hence, the origin is equidistant from the given lines.
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Similar questions
Q.
A point equidistant from the line 4x + 3y + 10 = 0, 5x − 12y + 26 = 0 and 7x+ 24y − 50 = 0 is
(a) (1, −1)
(b) (1, 1)
(c) (0, 0)
(d) (0, 1)
Q.
Find the equation. of the straight line equidistant from the straight lines 4x +3y+10=0 and 4x +3y+2=0.
Q.
The equation of the bisector of the lines 4x+3y-6=0 and 5x+12y+9=0 containing the origin is given by
.