Reduce the following equations into intercept form and find their intercepts on the axes.

(i) 3x + 2y – 12 = 0

(ii) 4x – 3y = 6

(iii) 3y + 2 = 0

Asked by Pragya Singh | 1 year ago |  71

##### Solution :-

(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 | 1 year ago

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