Question

Find an equation of the line that satisfies the given conditions.Through
(−1, −2);
perpendicular to the line
2x + 5y + 5 = 0

Answer 1

plz help

Answer 2

-5x+2y-1=0

Answer 3

i'll go slow, this kind of problems require a lot of reasoning.
we know this:
\[m _{1}=-\frac{ 1 }{ m _{2} }\]
All this says is, we need a slope, with inverted number in order for them to be perpendicular.

Moving on to the problem:
we have the ecuation:
\[2x+5y+5=0\]
Now, we want it in the form y=mx+b, so we can see the slope.
\[2x+5=-5y\]=
\[\frac{ 2x+5 }{ -5 }=y\]
ordering the terms:
\[y=-\frac{ 2x }{ 5 }-1\]
so, we can see that the slope of the line1 is -2/5 so the slope of the second one is 5/2

we have one of the points it intersects,
so we'll use the ecuation:
\[y-y _{1}=m(x-x _{1})\]
Let's replace the terms:
\[y+2=\frac{ 5 }{ 2 }(x+1)\]
So the ecuation of the line that satisfies that conditions is:
\[y=\frac{ 5 }{ 2 }x+\frac{ 1 }{ 2 }\]