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प्रश्न
A straight line is passing through the point A(1, 2) with slope `5/12`. Find points on the line which are 13 units away from A
उत्तर
Slope of the line m = tan θ = `5/12`
sin θ = `5/13`
cos θ = `12/13`
The parametric equation of the line passing through the point (1, 2) making angle θ with x-axis is
`(x - 1)/cos theta` = r
`(y - 2)/sin theta` = r
Any point on this line is
`(x - 1)/costheta` = r, `y - 2)/sintheta` = r
x – 1 = r cos θ, y – 2 = r sin θ
x = 1 + r cos θ, y = 2 + r sin θ
(1 + r cos θ, 2 + r sin θ) ......(1)
Where r is the distance of any point from A(1, 2) on the line.
To find the point which is 13 units away from A(1, 2) on the line.
Substitute r = ± 13, cos θ = `12/13`,
sin θ = `5/13` in equation (1)
Required point = `(1 +- 13 (12/13), 2 +- 13 (5/13))`
= (1 ± 12, 2 ± 5)
= (1 + 12, 2 + 5)
= (1 + 12, 2 + 5) or (1 – 12, 2 – 5)
= (13, 7) or (– 11, – 3)
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