Find $k$ so that the line through $(-2,3)$ and $(k, 6)$ is a) parallel to $4 y-3 x=1$; b) perpendicular to $3 y+5 x=2$. If the line is parallel to $4 y-3 x=1$, then $k=2$. If the line is perpendicular to $3 y+5 x=2$, then $k=$

Step 1 ：Given two points (-2,3) and (k,6), the slope of the line passing through these points is given by \(\frac{6 - 3}{k - (-2)} = \frac{3}{k + 2}\).

Step 2 ：For the line to be parallel to the line 4y - 3x = 1, their slopes must be equal. The slope of the line 4y - 3x = 1 can be found by rearranging it into the form y = mx + c, which gives us a slope of 0.75.

Step 3 ：Setting the two slopes equal to each other, we get \(\frac{3}{k + 2} = 0.75\). Solving this equation for k, we find that k = 2.

Step 4 ：For the line to be perpendicular to the line 3y + 5x = 2, their slopes must be negative reciprocals of each other. The slope of the line 3y + 5x = 2 can be found by rearranging it into the form y = mx + c, which gives us a slope of -1.6666666666666667.

Step 5 ：Setting the slope of our line equal to the negative reciprocal of the slope of the given line, we get \(\frac{3}{k + 2} = -1/(-1.6666666666666667)\). Solving this equation for k, we find that k = 3.

Step 6 ：Final Answer: If the line is parallel to 4y - 3x = 1, then k = \(\boxed{2}\). If the line is perpendicular to 3y + 5x = 2, then k = \(\boxed{3}\).