find the equation of the line which satisfy the given condition: Passing through the points (–1, 1) and (2,

find the equation of the line which satisfy the given condition: Passing through the points (–1, 1) and (2, – 4).

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

find the equation of the line which satisfy the given condition: Passing through the points (–1, 1) and (2, – 4).

Given:

Focuses \[\left( -\text{ }\mathbf{1},\text{ }\mathbf{1} \right)\text{ }\mathbf{and}\text{ }\left( \mathbf{2},\text{ }-\text{ }\mathbf{4} \right)\]

We realize that the condition of the line going through the focuses (\[\mathbf{x1},\text{ }\mathbf{y1})\text{ }\mathbf{and}\text{ }\left( \mathbf{x2},\text{ }\mathbf{y2} \right)\] is given by

\[\mathbf{y}\text{ }\text{ }\mathbf{1}\text{ }=\text{ }-\text{ }\mathbf{5}/\mathbf{3}\text{ }\left( \mathbf{x}\text{ }+\text{ }\mathbf{1} \right)\]

\[\mathbf{3}\text{ }\left( \mathbf{y}\text{ }\text{ }\mathbf{1} \right)\text{ }=\text{ }\left( -\text{ }\mathbf{5} \right)\text{ }\left( \mathbf{x}\text{ }+\text{ }\mathbf{1} \right)\]

\[\mathbf{3y}\text{ }\text{ }\mathbf{3}\text{ }=\text{ }-\text{ }\mathbf{5x}\text{ }\text{ }\mathbf{5}\]

\[\mathbf{3y}\text{ }\text{ }\mathbf{3}\text{ }+\text{ }\mathbf{5x}\text{ }+\text{ }\mathbf{5}\text{ }=\text{ }\mathbf{0}\]

\[\mathbf{5x}\text{ }+\text{ }\mathbf{3y}\text{ }+\text{ }\mathbf{2}\text{ }=\text{ }\mathbf{0}\]

∴ The condition of the line is \[\mathbf{5x}\text{ }+\text{ }\mathbf{3y}\text{ }+\text{ }\mathbf{2}\text{ }=\text{ }\mathbf{0}.\]

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.

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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