find the equation of the line which satisfy the given condition: Perpendicular distance from the origin is 5 units and the angle made by the perpendicular with the positive x-axis is 30o.

Class 11 | Exercise 10.2 | Maths | NCERT | Straight Lines

Given:

$\mathbf{p}\text{ }=\text{ }\mathbf{5}$

and

$\mathbf{\omega }\text{ }=\text{ }\mathbf{30}{}^\circ$

We realize that the condition of the line having typical distance p from the beginning and point ω which the ordinary makes with the positive heading of x-axis is given by

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

Subbing the qualities in the situation, we get

$\mathbf{x}\text{ }\mathbf{cos30}{}^\circ \text{ }+\text{ }\mathbf{y}\text{ }\mathbf{sin30}{}^\circ \text{ }=\text{ }\mathbf{5}$

$\mathbf{x}\left( \surd \mathbf{3}/\mathbf{2} \right)\text{ }+\text{ }\mathbf{y}\left( \text{ }\mathbf{1}/\mathbf{2}\text{ } \right)\text{ }=\text{ }\mathbf{5}$

$\surd \mathbf{3}\text{ }\mathbf{x}\text{ }+\text{ }\mathbf{y}\text{ }=\text{ }\mathbf{5}\left( \mathbf{2} \right)\text{ }=\text{ }\mathbf{10}~~~~~~~~$

$\surd \mathbf{3}\text{ }\mathbf{x}\text{ }+\text{ }\mathbf{y}\text{ }\text{ }\mathbf{-10}\text{ }=\text{ }\mathbf{0}$

∴ The condition of the line is

$\surd \mathbf{3}\text{ }\mathbf{x}\text{ }+\text{ }\mathbf{y}\text{ }\text{ }\mathbf{-10}\text{ }=\text{ }\mathbf{0}.$

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.

A line passes through the point (2, 1, -3) and is parallel to the vector Find the equations of the line in vector and Cartesian forms.

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.

Evaluate:

If $f(x) = |x| - 3,find\mathop {\lim }\limits_{x \to 3} f(x)$

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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More Material

Find the angle between each of the following pairs of lines:

Find the coordinates of the foot of the perpendicular drawn from the point (1, 2, 3) to the line

Show that the lines and do not intersect each other.

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.

A line passes through the point (2, 1, -3) and is parallel to the vector Find the equations of the line in vector and Cartesian forms.

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.

Evaluate:

Evaluate:

Evaluate:

If $f(x) = |x| - 3,find\mathop {\lim }\limits_{x \to 3} f(x)$

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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