Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x-axis. (i) x – √3y + 8 = 0 (ii) y

Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x-axis. (i) x – √3y + 8 = 0 (ii) y – 2 = 0

Class 11 | Exercise 10.3 | Maths | NCERT | Straight Lines

Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x-axis. (i) x – √3y + 8 = 0 (ii) y – 2 = 0

(i) x – √3y + 8 = 0

Given:

The condition is

$\mathbf{x}\text{ }\text{ }\surd \mathbf{3y}\text{ }+\text{ }\mathbf{8}\text{ }=\text{ }\mathbf{0}$

Condition of line in typical structure is given by

$\mathbf{x}\text{ }\mathbf{cos}\text{ }\mathbf{\theta }\text{ }+\text{ }\mathbf{y}\text{ }\mathbf{sin}\text{ }\mathbf{\theta }\text{ }=\text{ }\mathbf{p}$

where ‘θ’ is the point among opposite and positive x pivot and ‘p’ is opposite separation from beginning.

So presently,

$\mathbf{x}\text{ }\text{ }\surd \mathbf{3y}\text{ }+\text{ }\mathbf{8}\text{ }=\text{ }\mathbf{0}$

$\mathbf{x}\text{ }\text{ }\surd \mathbf{3y}\text{ }=\text{ }-\text{ }\mathbf{8}$

Separation both the sides by

$\surd \left( \mathbf{12}\text{ }+\text{ }\left( \surd \mathbf{3} \right)\mathbf{2} \right)\text{ }=\text{ }\surd \left( \mathbf{1}\text{ }+\text{ }\mathbf{3} \right)\text{ }=\text{ }\surd \mathbf{4}\text{ }=\text{ }\mathbf{2}$

$\mathbf{x}/\mathbf{2}\text{ }\text{ }\surd \mathbf{3y}/\mathbf{2}\text{ }=\text{ }-\text{ }\mathbf{8}/\mathbf{2}$

$\left( -\text{ }\mathbf{1}/\mathbf{2} \right)\mathbf{x}\text{ }+\text{ }\surd \mathbf{3}/\mathbf{2y}\text{ }=\text{ }\mathbf{4}$

This is as:

$\mathbf{x}\text{ }\mathbf{cos}\text{ }\mathbf{120o}\text{ }+\text{ }\mathbf{y}\text{ }\mathbf{sin}\text{ }\mathbf{120o}\text{ }=\text{ }\mathbf{4}$

∴ The above condition is of the structure, where

$\mathbf{\theta }\text{ }=\text{ }\mathbf{120}{}^\circ$

and

$\mathbf{p}\text{ }=\text{ }\mathbf{4}.$

Opposite distance of line from beginning

$=\text{ }\mathbf{4}$

Point among opposite and positive x –axis

$=\text{ }\mathbf{120}{}^\circ$

$\left( \mathbf{ii} \right)\text{ }\mathbf{y}\text{ }\text{ }\mathbf{2}\text{ }=\text{ }\mathbf{0}$

Given:

The condition is

$\mathbf{y}\text{ }\text{ }\mathbf{2}\text{ }=\text{ }\mathbf{0}$

Condition of line in typical structure is given by

$\mathbf{x}\text{ }\mathbf{cos}\text{ }\mathbf{\theta }\text{ }+\text{ }\mathbf{y}\text{ }\mathbf{sin}\text{ }\mathbf{\theta }\text{ }=\text{ }\mathbf{p}$

where ‘θ’ is the point among opposite and positive x pivot and ‘p’ is opposite separation from beginning.

So presently,

$\mathbf{0}\text{ }\times \text{ }\mathbf{x}\text{ }+\text{ }\mathbf{1}\text{ }\times \text{ }\mathbf{y}\text{ }=\text{ }\mathbf{2}$

Separation the two sides by

$\surd \left( \mathbf{02}\text{ }+\text{ }\mathbf{12} \right)\text{ }=\text{ }\surd \mathbf{1}\text{ }=\text{ }\mathbf{1}$

$\mathbf{0}\text{ }\left( \mathbf{x} \right)\text{ }+\text{ }\mathbf{1}\text{ }\left( \mathbf{y} \right)\text{ }=\text{ }\mathbf{2}$

This is as:

$\mathbf{x}\text{ }\mathbf{cos}\text{ }\mathbf{90o}\text{ }+\text{ }\mathbf{y}\text{ }\mathbf{sin}\text{ }\mathbf{90o}\text{ }=\text{ }\mathbf{2}$

∴ The above condition is of the structure

$\mathbf{x}\text{ }\mathbf{cos}\text{ }\mathbf{\theta }\text{ }+\text{ }\mathbf{y}\text{ }\mathbf{sin}\text{ }\mathbf{\theta }\text{ }=\text{ }\mathbf{p}$

, where

$\mathbf{\theta }\text{ }=\text{ }\mathbf{90}{}^\circ$

and

$\mathbf{p}\text{ }=\text{ }\mathbf{2}.$

Opposite distance of line from beginning