Gradient of a line The line L is a reflection of the line $2y + 3x =9$ in the $y-$ axis (I had to draw the graph on the grid previously) 
Find gradient of the line L 
How would I go about solving this? 
 A: Hints
When you reflect around $y$, how does this change the equation of the curve in question? Think about if the original curve has a point $(x,y)$, which point will this map to if you transform it by reflecting around $y$?
So what is the new equation of the line? And what is its gradient?
A: Let the equation of the line L: $y=mx+c$ then the point of intersection $\left(0, \frac{9}{2}\right)$ of the line: $3x+2y=9$ with the y-axis must also lie on the line L thus we have $$\frac{9}{2}=m(0)+c $$$$\implies c=\frac{9}{2}$$
Now, the point of intersection of the line: $3x+2y=9$ with the x-axis is $(3, 0)$  
Since, the origin is the mid-point of line joining the points of intersection of the lines: $y=mx+c$ & $3x+2y=9$ with the x-axis. Hence, the line L intersects x-axis at $(-3, 0)$ which satisfies the equation of line L as follows $$0=m(-3)+c$$$$\implies m=\frac{c}{3}=\frac{\frac{9}{2}}{3}=\frac{3}{2}$$ 
Hence the slope of line L is $\color{#0b4}{m=\frac{3}{2}}$ & the equation of line L: ${3x-2y+9=0}$
A: Hint:
the reflection around the $y$ axis change a point $P=(x,y)$ to a point $P'=(-x,y)$ so the equation of the line become:
$$
2y-3x=9
$$
