**(i) **3x + 2y – 12 = 0

Given:

The equation is 3x + 2y – 12 = 0

Equation of line in intercept form is given by

\( \dfrac{x}{a}+\dfrac{y}{b}=1\), where ‘a’ and ‘b’ are intercepts on x axis and y – axis respectively.

So, 3x + 2y = 12

now let us divide both sides by 12, we get

\( \dfrac{3x}{12}+\dfrac{2y}{12}=\dfrac{12}{12}\)

\( \dfrac{x}{4}+\dfrac{y}{6}=1\)

The above equation is of the form

\( \dfrac{x}{a}+\dfrac{y}{b}=1\), where a = 4, b = 6

Intercept on x – axis is 4

Intercept on y – axis is 6

**(ii) **4x – 3y = 6

Given:

The equation is 4x – 3y = 6

Equation of line in intercept form is given by

\( \dfrac{x}{a}+\dfrac{y}{b}=1\), where ‘a’ and ‘b’ are intercepts on x axis and y – axis respectively.

So, 4x – 3y = 6

Now let us divide both sides by 6, we get

\( \dfrac{4x}{6}-\dfrac{3y}{6}=\dfrac{6}{6}\)

\( \dfrac{2x}{3}-\dfrac{y}{2}=1\)

\( \dfrac{x}{\dfrac{3}{2}}+\dfrac{y}{-\dfrac{y}{2}}=1\)

The above equation is of the form

\( \dfrac{x}{a}+\dfrac{y}{b}=1\), where a = \( \dfrac{3}{2}\), b = -2

Intercept on x – axis is \( \dfrac{3}{2}\)

Intercept on y – axis is -2

**(iii) **3y + 2 = 0

Given:

The equation is 3y + 2 = 0

Equation of line in intercept form is given by

\( \dfrac{x}{a}+\dfrac{y}{b}=1\), where ‘a’ and ‘b’ are intercepts on x axis and y – axis respectively.

So, 3y = -2

Now, let us divide both sides by -2, we get

\( -\dfrac{3y}{2}-\dfrac{-2}{-2}=1\)

\( -\dfrac{3y}{2}=1\)

\( \dfrac{y}{-\dfrac{2}{3}}=1\)

The above equation is of the form

\( \dfrac{x}{a}+\dfrac{y}{b}=1\), where a = 0, b =\( - \dfrac{2}{3}\)

Intercept on x – axis is 0

Intercept on y – axis is \(- \dfrac{2}{3}\)

Answered by Abhisek | 11 months agoFind the angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1.

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Find the angles between pairs of straight lines 3x + y + 12 = 0 and x + 2y – 1 = 0