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प्रश्न
Reduce the following equation into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x-axis.
`x – sqrt3y + 8 = 0`
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उत्तर
The given equation is `"x" - sqrt3"y" + 8 = 0`
It can be reduced as:
`"x" - sqrt3"y" = -8`
⇒ `-"x" + sqrt3"y" = 8`
On dividing both sides by `sqrt((-1)^2 + (sqrt3)^2) = sqrt4 = 2` we obtain
`-"x"/2 + sqrt3/2"y" = 8/2`
⇒ `(-1/2)"x" + (sqrt3/2)"y" = 4`
⇒ x cos 120° + y sin 120° = 4 ..........(i)
Equation (i) is in the normal form.
On comparing equation (1) with the normal form of equation of line
x cos ω + y sin ω = p, we obtain ω = 120° and p = 4.
Thus, the perpendicular distance of the line from the origin is 4, while the angle between the perpendicular and the positive x-axis is 120°.
