Find the angle between the x-axis and the line joining the points (3, –1) and (4, –2).

Class 11 | Exercise 10.1 | Maths | NCERT | Straight Lines

Slope of the line joining the focuses \[\left( \mathbf{3},\text{ }-\text{ }\mathbf{1} \right)\text{ }\mathbf{and}\text{ }\left( \mathbf{4},\text{ }-\text{ }\mathbf{2} \right)\] is given by

\[\mathbf{m}\text{ }=\text{ }\left( \mathbf{y2}\text{ }\text{ }\mathbf{y1} \right)/\left( \mathbf{x2}\text{ }\text{ }\mathbf{x1} \right)\] where, \[\mathbf{x}\text{ }\ne \text{ }\mathbf{x1}\]

\[\mathbf{m}\text{ }=\text{ }\left( -\text{ }\mathbf{2}\text{ }\text{ }\left( -\text{ }\mathbf{1} \right) \right)/\left( \mathbf{4}-\mathbf{3} \right)\]

\[=\text{ }\left( -\text{ }\mathbf{2}+\mathbf{1} \right)/\left( \mathbf{4}-\mathbf{3} \right)\]

\[=\text{ }-\text{ }\mathbf{1}/\mathbf{1}\]

\[=\text{ }-\text{ }\mathbf{1}\]

point of tendency of line joining the focuses \[\left( \mathbf{3},\text{ }-\text{ }\mathbf{1} \right)\text{ }\mathbf{and}\text{ }\left( \mathbf{4},\text{ }-\text{ }\mathbf{2} \right)\] is given by

\[\mathbf{tan}\text{ }\mathbf{\theta }\text{ }=\text{ }-\text{ }\mathbf{1}\]

\[\mathbf{\theta }\text{ }=\text{ }\left( \mathbf{90}{}^\circ \text{ }+\text{ }\mathbf{45}{}^\circ \right)\text{ }=\text{ }\mathbf{135}{}^\circ \]

∴ The point between the x-pivot and the line joining the focuses \[\left( \mathbf{3},\text{ }\text{ }\mathbf{1} \right)\text{ }\mathbf{and}\text{ }\left( \mathbf{4},\text{ }\text{ }\mathbf{2} \right)\] is \[\mathbf{135}{}^\circ .\]

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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.

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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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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