If three lines whose equations are y = m1x + c1, y = m2x + c2 and y = m3x + c3 are concurrent, then show that m1 (c2 – c3) + m2 (c3 – c1) + m3 (c1

If three lines whose equations are y = m1x + c1, y = m2x + c2 and y = m3x + c3 are concurrent, then show that m1 (c2 – c3) + m2 (c3 – c1) + m3 (c1 – c2) = 0

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

If three lines whose equations are y = m1x + c1, y = m2x + c2 and y = m3x + c3 are concurrent, then show that m1 (c2 – c3) + m2 (c3 – c1) + m3 (c1 – c2) = 0

It is provided that

$y\text{ }=\text{ }{{m}_{1}}x\text{ }+\text{ }{{c}_{1}}~\ldots ..\text{ }\left( 1 \right)$

$y\text{ }=\text{ }m2x\text{ }+\text{ }c2\text{ }\ldots \text{ }..\text{ }\left( 2 \right)$

$y\text{ }=\text{ }m3x\text{ }+\text{ }c3\text{ }\ldots \text{ }..\text{ }\left( 3 \right)$

By deducting condition (1) from (2) we get

$0\text{ }=\text{ }\left( m2\text{ }\text{ }m1 \right)\text{ }x\text{ }+\text{ }\left( c2\text{ }\text{ }c1 \right)$

$\left( m1\text{ }\text{ }m2 \right)\text{ }x\text{ }=\text{ }c2\text{ }\text{ }c1$

So we get

NCERT Solutions for Class 11 Maths Chapter 10 – Straight Lines image - 56

NCERT Solutions for Class 11 Maths Chapter 10 – Straight Lines image - 57

Taking out the normal terms

$m1\text{ }\left( c2\text{ }\text{ }c3 \right)\text{ }+\text{ }m2\text{ }\left( c3\text{ }\text{ }c1 \right)\text{ }+\text{ }m3\text{ }\left( c1\text{ }\text{ }c2 \right)\text{ }=\text{ }0$

Thus,

$m1\text{ }\left( c2\text{ }\text{ }c3 \right)\text{ }+\text{ }m2\text{ }\left( c3\text{ }\text{ }c1 \right)\text{ }+\text{ }m3\text{ }\left( c1\text{ }\text{ }c2 \right)\text{ }=\text{ }0.$

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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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.

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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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