Question -
Answer -
(i) 3x + 2y – 12 = 0
Given:
The equation is 3x + 2y – 12 = 0
Equation of line in intercept form is given by x/a + 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
3x/12 + 2y/12 = 12/12
x/4 + y/6 = 1
∴ The above equation is of the form x/a + 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 x/a + 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
4x/6 – 3y/6 = 6/6
2x/3 – y/2 = 1
x/(3/2) + y/(-2) = 1
∴ The above equation is of the form x/a + y/b = 1, where a = 3/2, b = -2
Intercept on x – axis is 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 x/a + 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
3y/-2 = -2/-2
3y/-2 = 1
y/(-2/3) = 1
∴ The above equation is of the form x/a + y/b = 1, where a = 0, b = -2/3
Intercept on x – axis is 0
Intercept on y – axis is -2/3