Question

Find the equation of the line that contains the point (-4, -4) and is perpendicular to the line 4x+5y=5

Answer 1

Answer:

$5x-4y+4=0$

Step-by-step explanation:

Given: A point  which is perpendicular to the line .

To find: The equation of the line which passes through $(-4, -4)$ and is perpendicular to the line $4x+5y=5$.

Solution:

We have, $4x+5y=5$.

Slope of the line $4x+5y=5$ is $-\frac{4}{5}$.

The line which passes through $(-4, -4)$ is perpendicular to the line $4x+5y=5$.

Now, the product of the slopes of two perpendicular lines is $-1$.

Therefore, $m\times-\frac{4}{5}=-1$

So, its slope is $\frac{5}{4}$.

Now, the equation of the line which passes through $(-4, -4)$ and slope $\frac{5}{4}$ is:

$y-(-4)=\frac{5}{4} [x-(-4)]$

$\Rightarrow y+4=\frac{5}{4} (x+4)$

$\Rightarrow 4(y+4)=5(x+4)$

$\Rightarrow 4y+16=5x+20$

$\Rightarrow 5x-4y+20-16=0$

$\Rightarrow 5x-4y+4=0$

Hence, the equation of the line that contains the point $(-4, -4)$ and is perpendicular to the line $4x+5y=5$ is $5x-4y+4=0$.