Question

transform the following equation √3x+y+10=0 into 1)normal form,2)intercept form and 3)slope intercept form​

Answer 1

SOLUTION

TO DETERMINE

To transform the following equation √3x+y+10=0 into

1)normal form

2)intercept form

3)slope intercept form

CONCEPT TO BE IMPLEMENTED

General form

The general form of any line is ax + by = c

Normal form

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

Slope - Intercept form

y = mx + c

Intercept form

$\displaystyle \sf{ \frac{x}{a} + \frac{y}{b} = 1}$

EVALUATION

Here the given equation of the line is

$\sf{ \sqrt{3}x + y + 10 = 0 }$

1. Normal form

Here the given equation of the line is

$\sf{ \sqrt{3}x + y + 10 = 0 }$

The given equation of the line can be rewritten as

$\displaystyle \sf{ - \frac{ \sqrt{3} x}{ \sqrt{3 + 1} } - \frac{y}{ \sqrt{3 + 1} } = \frac{10}{ \sqrt{3 + 1} } }$

$\displaystyle \sf{ \implies \: - \frac{ \sqrt{3} }{2 }x - \frac{y}{2} = 5 }$

$\displaystyle \sf{ \implies \: x \cos \frac{7\pi}{6} + y \sin \frac{7\pi}{6} = 5 }$

Which is of the form

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

2. Intercept form

Here the given equation of the line is

$\sf{ \sqrt{3}x + y + 10 = 0 }$

Which can be rewritten as

$\displaystyle \sf{ \implies \frac{x}{\displaystyle \sf{ - \frac{10}{ \sqrt{3} } } } + \frac{y}{ - 10} = 1 }$

Which is of the intercept form

3. Slope intercept form

Here the given equation of the line is

$\sf{ \sqrt{3}x + y + 10 = 0 }$

Which can be rewritten as

$\sf{y = - \sqrt{3} x - 10}$

Which is of the intercept form

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Answer 2

Answer:

Step-by-step explanation: