Question -
Answer -
(i) 3x + 1/7y = 3
7x + 3y = 7
(ii) –2x – 3y = 1
6y + 4x = – 2
(iii) x/2 + y + 2/5 =0
4x + 8y + 5/16 = 0
Solution:
Condition forcoincident lines,
a1/a2 =b1/b2 = c1/c2;
(i) No.
Given pair of linearequations are:
3x + 1/7y = 3
7x + 3y = 7
Comparing the aboveequations with ax + by + c = 0;
Here, a1 =3, b1 = 1/7, c1 = – 3;
And a2 =7, b2 = 3, c2 = – 7;
a1 /a2 =3/7
b1 /b2 =1/21
c1 /c2 =– 3/ – 7 = 3/7
Here, a1/a2 ≠b1/b2.
Hence, the given pairof linear equations has unique solution.
(ii) Yes,
Given pair of linearequations.
– 2x – 3y – 1 = 0 and4x + 6y + 2 = 0;
Comparing the aboveequations with ax + by + c = 0;
Here, a1 =– 2, b1 = – 3, c1 = – 1;
And a2 =4, b2 = 6, c2 = 2;
a1 /a2 =– 2/4 = – ½
b1 /b2 =– 3/6 = – ½
c1 /c2 =– ½
Here, a1/a2 =b1/b2 = c1/c2, i.e. coincidentlines
Hence, the given pairof linear equations is coincident.
(iii) No,
Given pair of linearequations are
x/2 + y + 2/5 = 0
4x + 8y + 5/16 = 0
Comparing the aboveequations with ax + by + c = 0;
Here, a1 =½, b1 = 1, c1 = 2/5;
And a2 =4, b2 = 8, c2 = 5/16;
a1 /a2 =1/8
b1 /b2 =1/8
c1 /c2 =32/25
Here, a1/a2 =b1/b2 ≠ c1/c2, i.e.parallel lines
Hence, the given pairof linear equations has no solution.