Question

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}

Answer 1

Answer:

y = - $\frac{4}{5}$ x + $\frac{2}{5}$

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 = $\frac{5}{4}$ x + $\frac{3}{4}$ ← in slope- intercept form

with slope m = $\frac{5}{4}$

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

$m_{perpendicular}$ = - $\frac{1}{m}$ = - $\frac{1}{\frac{5}{4} }$ = - $\frac{4}{5}$ , thus

y = - $\frac{4}{5}$ x + c ← is the partial equation

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

- 2 = - $\frac{12}{5}$ + c ⇒ c = - 2 + $\frac{12}{5}$ = $\frac{2}{5}$

y = - $\frac{4}{5}$ x + $\frac{2}{5}$ ← equation of perpendicular line