By using the concept of equation of a line, prove that the three points (3, 0), (– 2, – 2) and (8, 2) are collinear

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

On the off chance that we need to demonstrate that the given three focuses

$\left( \mathbf{3},\text{ }\mathbf{0} \right),\text{ }\left( \text{ }\mathbf{2},\text{ }\text{ }\mathbf{2} \right)\text{ }\mathbf{and}\text{ }\left( \mathbf{8},\text{ }\mathbf{2} \right)$

are collinear, then, at that point, we need to likewise demonstrate that the line going through the focuses

$\left( \mathbf{3},\text{ }\mathbf{0} \right)\text{ }\mathbf{and}\text{ }\left( \text{ }\mathbf{2},\text{ }\text{ }\mathbf{2} \right)$

additionally goes through the point

$\left( \mathbf{8},\text{ }\mathbf{2} \right).$

By utilizing the equation,

The condition of the line going through the focuses

$\left( \mathbf{x1},\text{ }\mathbf{y1} \right)\text{ }\mathbf{and}\text{ }\left( \mathbf{x2},\text{ }\mathbf{y2} \right)$

is given by

$-\text{ }\mathbf{5y}\text{ }=\text{ }-\text{ }\mathbf{2}\text{ }\left( \mathbf{x}\text{ }-\text{ }\mathbf{3} \right)$

$-\text{ }\mathbf{5y}\text{ }=\text{ }-\text{ }\mathbf{2x}\text{ }+\text{ }\mathbf{6}$

$\mathbf{2x}\text{ }\text{ }\mathbf{-5y}\text{ }=\text{ }\mathbf{6}$

On the off chance that

$\mathbf{2x}\text{ }\text{ }\mathbf{-5y}\text{ }=\text{ }\mathbf{6}$

goes through

$\left( \mathbf{8},\text{ }\mathbf{2} \right),$

$\mathbf{2x}\text{ }\text{ }\mathbf{-5y}\text{ }=\text{ }\mathbf{2}\left( \mathbf{8} \right)\text{ }\text{ }\mathbf{-5}\left( \mathbf{2} \right)$

$=\text{ }\mathbf{16}\text{ }-\text{ }\mathbf{10}$

$=\text{ }\mathbf{6}$

$=\text{ }\mathbf{RHS}$

The line going through the focuses

$\left( \mathbf{3},\text{ }\mathbf{0} \right)\text{ }\mathbf{and}\text{ }\left( \text{ }\mathbf{2},\text{ }\text{ }\mathbf{2} \right)$

likewise goes through the point

$\left( \mathbf{8},\text{ }\mathbf{2} \right).$

Thus demonstrated. The given three focuses are collinear.

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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