The base of an equilateral triangle with side 2a lies along the y-axis such that the mid-point of the base is at the origin. Find vertices of the triangle

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

NCERT Solutions for Class 11 Maths Chapter 10 – Straight Lines image - 2

Allow us to consider ABC be the given symmetrical triangle with side 2a.

Where,

$\mathbf{AB}\text{ }=\text{ }\mathbf{BC}\text{ }=\text{ }\mathbf{AC}\text{ }=\text{ }\mathbf{2a}$

In the above figure, by accepting that the base BC lies on the x hub with the end goal that the mid-point of BC is at the beginning for example

$\mathbf{BO}\text{ }=\text{ }\mathbf{OC}\text{ }=\text{ }\mathbf{a}$

, where O is the beginning.

The co-ordinates of point C are (0, a) and that of B are (0,- a)

Since the line joining a vertex of a symmetrical ∆ with the mid-point of its contrary side is opposite.

Along these lines, vertex A lies on the y – hub

By applying Pythagoras hypothesis

$\left( \mathbf{AC} \right)\mathbf{2}\text{ }=\text{ }\mathbf{OA2}\text{ }+\text{ }\mathbf{OC2}$

$\left( \mathbf{2a} \right)\mathbf{2}=\text{ }\mathbf{a2}\text{ }+\text{ }\mathbf{OC2}$

$\mathbf{4a2}\text{ }\text{ }\mathbf{a2}\text{ }=\text{ }\mathbf{OC2}$

$\mathbf{3a2}\text{ }=\text{ }\mathbf{OC2}$

$\mathbf{OC}\text{ }=\surd \mathbf{3a}$

Co-ordinates of point C =

$\pm \text{ }\surd \mathbf{3a},\text{ }\mathbf{0}$

∴ The vertices of the given symmetrical triangle are

$\left( \mathbf{0},\text{ }\mathbf{a} \right),\text{ }\left( \mathbf{0},\text{ }-\text{ }\mathbf{a} \right),\text{ }\left( \surd \mathbf{3a},\text{ }\mathbf{0} \right)$

Or on the other hand

$\left( \mathbf{0},\text{ }\mathbf{a} \right),\text{ }\left( \mathbf{0},\text{ }-\text{ }\mathbf{a} \right)\text{ }\mathbf{and}\text{ }\left( -\text{ }\surd \mathbf{3a},\text{ }\mathbf{0} \right)$

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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