Question

Transform the following equation 3 10 0 x y    in to
(a) Slope – intercept form
(b) intercept form and (c) normal from.

Answer 1

Answer:

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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