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the straight line L has the equation 4y=5x+3. Point A has the coordinates (3,-2) Find the equation of the line straight line that is perpendicular to L and passes through A. Write your answer in the form y-mx+c where m and c are decimals or fractions {3 marks}

Respuesta :

Answer:

y = - [tex]\frac{4}{5}[/tex] x + [tex]\frac{2}{5}[/tex]

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Rearrange 4y = 5x + 3 into this form by dividing the 3 terms by 4

y = [tex]\frac{5}{4}[/tex] x + [tex]\frac{3}{4}[/tex] ← in slope- intercept form

with slope m = [tex]\frac{5}{4}[/tex]

Given a line with slope m then the slope of a line perpendicular to it is

[tex]m_{perpendicular}[/tex] = - [tex]\frac{1}{m}[/tex] = - [tex]\frac{1}{\frac{5}{4} }[/tex] = - [tex]\frac{4}{5}[/tex] , thus

y = - [tex]\frac{4}{5}[/tex] x + c ← is the partial equation

To find c substitute (3, - 2) into the partial equation

- 2 = - [tex]\frac{12}{5}[/tex] + c ⇒ c = - 2 + [tex]\frac{12}{5}[/tex] = [tex]\frac{2}{5}[/tex]

y = - [tex]\frac{4}{5}[/tex] x + [tex]\frac{2}{5}[/tex] ← equation of perpendicular line