Find the equation of the line parallel to y-axis and drawn through the point of intersection of the lines x – 7y + 5 = 0 and 3x + y = 0.

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

Here the situation of any line corresponding to the y-pivot is of the structure

$\mathbf{x}\text{ }=\text{ }\mathbf{a}\text{ }\ldots \text{ }.\text{ }\left( \mathbf{1} \right)$

Two given lines are

$\mathbf{x}\text{ }\text{ }\mathbf{7y}\text{ }+\text{ }\mathbf{5}\text{ }=\text{ }\mathbf{0}\text{ }\ldots \text{ }\left( \mathbf{2} \right)$

$\mathbf{3x}\text{ }+\text{ }\mathbf{y}\text{ }=\text{ }\mathbf{0}\text{ }\ldots \text{ }\left( \mathbf{3} \right)$

By addressing conditions (2) and (3) we get

$\mathbf{x}\text{ }=\text{ }-\text{ }\mathbf{5}/\mathbf{22}\text{ }\mathbf{and}\text{ }\mathbf{y}\text{ }=\text{ }\mathbf{15}/\mathbf{22}$

$\left( -\text{ }\mathbf{5}/\mathbf{22},\text{ }\mathbf{15}/\mathbf{22} \right)$

is the mark of convergence of lines (2) and (3)

In the event that the line

$\mathbf{x}\text{ }=\text{ }\mathbf{a}$

goes through point

$\left( -\text{ }\mathbf{5}/\mathbf{22},\text{ }\mathbf{15}/\mathbf{22} \right)$

we get

$\mathbf{a}\text{ }=\text{ }-\text{ }\mathbf{5}/\mathbf{22}$

Consequently, the necessary condition of the line is

$\mathbf{x}\text{ }=\text{ }-\text{ }\mathbf{5}/\mathbf{22}.$

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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