Question

Write an equation of the line that passes through the given point and is perpendicular to the given equation : (-3,-2); 2x+4y=7 and (4,-3); y= 3/7x-5

Answer 1

Please help!

Answer 2

Normal vector of 2x+4y=7 is \[\left(\begin{matrix}2 \\ 4\end{matrix}\right)\]Then a normal vector of lines perpendicular to it is:\[\left(\begin{matrix}-4 \\ 2\end{matrix}\right)\]leading to the equation -4x+2y=c. If you substitute x=-3 and y=-2 in this equation, you get c.

Answer 3

1. The slope for the new equation would be 2x becaus eof the recipical
Y-Y1= m(X-X1)
Y-(-2)=2(X+3)
y+2=2x+6
y=2x+4
2. 7/3x is the new slope
Y-Y1= m(x-x1)
Y-(-3)=-7/3(x-4)
Y+3=-7/3x+9 1/3
y=-7/3x+6 1/3

Answer 4

^ its wrong

Answer 5

x=-3, y=-2:
-4*-3+2*-2=12-4=8,
so eq is -4x+2y=8, or (simpler)

-2x+y=4