find the equation of the line which satisfy the given condition: Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9

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

We realize that condition of the line making blocks an and b on x-and y-axis, individually, is

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

Given: amount of captures = 9

$\mathbf{a}\text{ }+\text{ }\mathbf{b}\text{ }=\text{ }\mathbf{9}$

$\mathbf{b}\text{ }=\text{ }\mathbf{9}\text{ }\text{ }\mathbf{a}$

Presently, substitute worth of b in the above condition, we get

$\mathbf{x}/\mathbf{a}\text{ }+\text{ }\mathbf{y}/\left( \mathbf{9}\text{ }\text{ }\mathbf{a} \right)\text{ }=\text{ }\mathbf{1}$

Given: the line goes through the point (2, 2),

In this way,

$\mathbf{2}/\mathbf{a}\text{ }+\text{ }\mathbf{2}/\left( \mathbf{9}\text{ }\text{ }\mathbf{a} \right)\text{ }=\text{ }\mathbf{1}$

$\left[ \mathbf{2}\left( \mathbf{9}\text{ }\text{ }\mathbf{a} \right)\text{ }+\text{ }\mathbf{2a} \right]/\mathbf{a}\left( \mathbf{9}\text{ }\text{ }\mathbf{a} \right)\text{ }=\text{ }\mathbf{1}$

$\left[ \mathbf{18}\text{ }\text{ }\mathbf{2a}\text{ }+\text{ }\mathbf{2a} \right]/\mathbf{a}\left( \mathbf{9}\text{ }\text{ }\mathbf{a} \right)\text{ }=\text{ }\mathbf{1}$

$\mathbf{18}/\mathbf{a}\left( \mathbf{9}\text{ }\text{ }\mathbf{a} \right)\text{ }=\text{ }\mathbf{1}$

$\mathbf{18}\text{ }=\text{ }\mathbf{a}\text{ }\left( \mathbf{9}\text{ }\text{ }\mathbf{a} \right)$

$\mathbf{18}\text{ }=\text{ }\mathbf{9a}\text{ }\text{ }\mathbf{a2}$

$\mathbf{a2}\text{ }\text{ }\mathbf{9a}\text{ }+\text{ }\mathbf{18}\text{ }=\text{ }\mathbf{0}$

After factorizing, we get

$\mathbf{a2}\text{ }\text{ }\mathbf{3a}\text{ }\text{ }\mathbf{6a}\text{ }+\text{ }\mathbf{18}\text{ }=\text{ }\mathbf{0}$

$\left( \mathbf{a}\text{ }\text{ }\mathbf{3} \right)\text{ }\text{ }\mathbf{6}\text{ }\left( \mathbf{a}\text{ }\text{ }\mathbf{3} \right)\text{ }=\text{ }\mathbf{0}$

$\left( \mathbf{a}\text{ }\text{ }\mathbf{3} \right)\text{ }\left( \mathbf{a}\text{ }\text{ }\mathbf{6} \right)\text{ }=\text{ }\mathbf{0}$

$\mathbf{a}\text{ }=\text{ }\mathbf{3}\text{ }\mathbf{or}\text{ }\mathbf{a}\text{ }=\text{ }\mathbf{6}$

Allow us to substitute in (1),

$\mathbf{Case}\text{ }\mathbf{1}\text{ }\left( \mathbf{a}\text{ }=\text{ }\mathbf{3} \right):$

Then, at that point,

$\mathbf{b}\text{ }=\text{ }\mathbf{9}\text{ }\text{ }\mathbf{3}\text{ }=\text{ }\mathbf{6}$

$\mathbf{x}/\mathbf{3}\text{ }+\text{ }\mathbf{y}/\mathbf{6}\text{ }=\text{ }\mathbf{1}$

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

$\mathbf{2x}\text{ }+\text{ }\mathbf{y}\text{ }\text{ }\mathbf{6}\text{ }=\text{ }\mathbf{0}$

$\mathbf{Case}\text{ }\mathbf{2}\text{ }\left( \mathbf{a}\text{ }=\text{ }\mathbf{6} \right):$

Then, at that point,

$\mathbf{b}\text{ }=\text{ }\mathbf{9}\text{ }\text{ }\mathbf{6}\text{ }=\text{ }\mathbf{3}$

$\mathbf{x}/\mathbf{6}\text{ }+\text{ }\mathbf{y}/\mathbf{3}\text{ }=\text{ }\mathbf{1}$

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

$\mathbf{x}\text{ }+\text{ }\mathbf{2y}\text{ }\text{ }\mathbf{6}\text{ }=\text{ }\mathbf{0}$

∴ The condition of the line is

$\mathbf{2x}\text{ }+\text{ }\mathbf{y}\text{ }\text{ }\mathbf{6}\text{ }=\text{ }\mathbf{0}\text{ }\mathbf{or}\text{ }\mathbf{x}\text{ }+\text{ }\mathbf{2y}\text{ }\text{ }\mathbf{6}\text{ }=\text{ }\mathbf{0}.$

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.

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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