Question

transform the following equation 4x - 3y +12 =0 into .slope intercept , intercept form and normal form​

Answer 1

SOLUTION

TO TRANSFORM

The equation 4x - 3y +12 =0 into

• Slope intercept form

• Intercept form

• Normal form

EVALUATION

Transform Into Slope intercept form

Here the given equation is 4x - 3y +12 = 0

Which can be rewritten as below :

$\sf{3y = 4x + 12}$

$\implies \displaystyle \sf{y = \frac{4}{3} x + 4}$

Which is of the Slope Intercept form

Transform Into Intercept form

Here the given equation is 4x - 3y + 12 = 0

Which can be rewritten as below :

$\displaystyle \sf{4x - 3y = - 12}$

$\implies\displaystyle \sf{ \frac{4x}{ - 12} + \frac{ - 3y}{ - 12} = 1 }$

$\implies\displaystyle \sf{ \frac{x}{ - 3} + \frac{ y}{ 4} = 1 }$

Which is of the intercept form

Transform Into Normal form

Here the given equation is 4x - 3y + 12 = 0

Which can be rewritten as below :

$\displaystyle \sf{4x - 3y = - 12 }$

$\implies\displaystyle \sf{ \frac{4x}{ \sqrt{ {4}^{2} + {3}^{2} } } + \frac{ - 3y}{ \sqrt{ {4}^{2} + {3}^{2} } } = \frac{ - 12}{ \sqrt{ {4}^{2} + {3}^{2} } } }$

$\implies \sf{x \cos \alpha + y \sin \alpha = p}$

Which is of the Normal form

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Answer 2

Answer:

wkt slope intercept form y=mx+c

and intercept form x/a+y/b=1