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Sagot :
Given:
'a' and 'b' are the intercepts made by a straight-line with the co- ordinate axes.
3a = b and the line pass through the point (1, 3).
To find:
The equation of the line.
Solution:
The intercept form of a line is
[tex]\dfrac{x}{a}+\dfrac{y}{b}=1[/tex] ...(i)
where, a is x-intercept and b is y-intercept.
We have, 3a=b.
[tex]\dfrac{x}{a}+\dfrac{y}{3a}=1[/tex] ...(ii)
The line pass through the point (1, 3). So, putting x=1 and y=3, we get
[tex]\dfrac{1}{a}+\dfrac{3}{3a}=1[/tex]
[tex]\dfrac{1}{a}+\dfrac{1}{a}=1[/tex]
[tex]\dfrac{2}{a}=1[/tex]
Multiply both sides by a.
[tex]2=a[/tex]
The value of a is 2. So, x-intercept is 2.
Putting a=2 in [tex]b=3a[/tex], we get
[tex]b=3(2)[/tex]
[tex]b=6[/tex]
The value of b is 6. So, y-intercept is 6.
Putting a=2 and b=6 in (i), we get
[tex]\dfrac{x}{2}+\dfrac{y}{6}=1[/tex]
Therefore, the equation of the required line in intercept form is [tex]\dfrac{x}{2}+\dfrac{y}{6}=1[/tex].
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