equation of the line passing through (0,2)and having intercepts in the ratio 2:3 is
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Let the equation of the line be
x/a + y/b = 1 ........ (1)
which shows that the line meets the coordinate axis at A(a,0) and B(0,b) respectively. The coordinates of the points divides the line by the ratio 2:3
Therefore,
( (2.0 + 3.a) / 2+3 , (2.b + 3.0) / 2+3 ) [ ∵ (mx₂+nx₁/m+n,my₂+ny₁/m+n) ]
(3a/5 , 2b/5)
Given point is (0,2)
Hence,
3a/5 = 0 , 2b/5 = 2
a = 0 , b = 5
Put these values in equation (1), we get:
x/0 + y/5 = 1
0 + y/5 = 1
y/5 = 1
y = 5
y - 5 = 0
which is the required equation.
Hope it will help you. Thanks.
x/a + y/b = 1 ........ (1)
which shows that the line meets the coordinate axis at A(a,0) and B(0,b) respectively. The coordinates of the points divides the line by the ratio 2:3
Therefore,
( (2.0 + 3.a) / 2+3 , (2.b + 3.0) / 2+3 ) [ ∵ (mx₂+nx₁/m+n,my₂+ny₁/m+n) ]
(3a/5 , 2b/5)
Given point is (0,2)
Hence,
3a/5 = 0 , 2b/5 = 2
a = 0 , b = 5
Put these values in equation (1), we get:
x/0 + y/5 = 1
0 + y/5 = 1
y/5 = 1
y = 5
y - 5 = 0
which is the required equation.
Hope it will help you. Thanks.
Answered by
115
Answer:
equation of the line passing through (0,2)and having intercepts in the ratio 2:3 is
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