Find 'a" so that (3,a) lies on the line represented by 2x-3y-5=0. Also , find the coordinates of the point where line cuts the x-axis.
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Answered by
27
It's a good question....
Since the pt 3,a lies on the line, the pt should satisfy the eqn of the line.
ie. X=3, y=a
2×3-3× a -5=0
6-3a-5=0
1-3a=0
3a=1
a=1/3
____
Now, the line cuts the x axis means y=0.
2x-3×0-5=0
2x-5=0
x=5/2
Thus the pt is (5/2,0)
Since the pt 3,a lies on the line, the pt should satisfy the eqn of the line.
ie. X=3, y=a
2×3-3× a -5=0
6-3a-5=0
1-3a=0
3a=1
a=1/3
____
Now, the line cuts the x axis means y=0.
2x-3×0-5=0
2x-5=0
x=5/2
Thus the pt is (5/2,0)
Answered by
12
(3,a) lies on the line 2x-3y-5=0
so 2x-3y-5=0
2x-3y=5
2*3-3a=5
6-5=3a
1/3=a
on x axis every point is of the form(x,0)
so 2x-3y=5
2x-3*0=5
2x=5
x=5/2
so 2x-3y-5=0
2x-3y=5
2*3-3a=5
6-5=3a
1/3=a
on x axis every point is of the form(x,0)
so 2x-3y=5
2x-3*0=5
2x=5
x=5/2
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