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 \[=\text{ }\mathbf{2}\]

Point among opposite and positive x –axis \[=\text{ }\mathbf{90}{}^\circ \]

i

Find the vector equation of the line passing through the point with position vector and parallel to the vector Deduce the Cartesian equations of the line.

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A line passes through the point (3, 4, 5) and is parallel to the vector Find the equations of the line in the vector as well as Cartesian forms.

Let $\begin{array}{l} f(x) = \left\{ \begin{array}{l} \frac{x}{{|x|}},x \ne 0\\ 0,x = 0 \end{array} \right.\\ \end{array}$ . Show that $\begin{array}{l} \mathop {\lim }\limits_{x \to 0} f(x) \end{array}$ does not exist.

Let $\begin{array}{l} f(x) = \left\{ \begin{array}{l} \frac{|x-3|}{{(x-3)’}},x \ne 3\\ 0,x = 3 \end{array} \right.\\ \end{array}$ . Show that $\begin{array}{l} \mathop {\lim }\limits_{x \to 3} f(x) \end{array}$ does not exist.

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