Respuesta :
Answer:
[tex]5x-4y+4=0[/tex]
Step-by-step explanation:
Given: A point which is perpendicular to the line .
To find: The equation of the line which passes through [tex](-4, -4)[/tex] and is perpendicular to the line [tex]4x+5y=5[/tex].
Solution:
We have, [tex]4x+5y=5[/tex].
Slope of the line [tex]4x+5y=5[/tex] is [tex]-\frac{4}{5}[/tex].
The line which passes through [tex](-4, -4)[/tex] is perpendicular to the line [tex]4x+5y=5[/tex].
Now, the product of the slopes of two perpendicular lines is [tex]-1[/tex].
Therefore, [tex]m\times-\frac{4}{5}=-1[/tex]
So, its slope is [tex]\frac{5}{4}[/tex].
Now, the equation of the line which passes through [tex](-4, -4)[/tex] and slope [tex]\frac{5}{4}[/tex] is:
[tex]y-(-4)=\frac{5}{4} [x-(-4)][/tex]
[tex]\Rightarrow y+4=\frac{5}{4} (x+4)[/tex]
[tex]\Rightarrow 4(y+4)=5(x+4)[/tex]
[tex]\Rightarrow 4y+16=5x+20[/tex]
[tex]\Rightarrow 5x-4y+20-16=0[/tex]
[tex]\Rightarrow 5x-4y+4=0[/tex]
Hence, the equation of the line that contains the point [tex](-4, -4)[/tex] and is perpendicular to the line [tex]4x+5y=5[/tex] is [tex]5x-4y+4=0[/tex].
