1)
\(y=mx+b\)
where m is the slope
We know the slope is 5 and the line passes through the point (4,3).
\(m=5 \\
(4,3) \\
x=4 \\ y=3 \\ \\
3=5 \times 4 + b \\
3=20 + b \\
b=3-20 \\
b=-17\)
The equation of the line is:
\(y=5m-17\)
Now the second point:
\((2,y) \\
x=2 \\
y \\ \\ y=5 \times 2 -17 \\ y=10-17 \\ y=-7\)
The answer:
The value of y is -7.
2)
The line passes through the origin, or the point (0,0), and has a slope of -32.
The slope is negative so the line goes down from left to right.
It crosses both x and y axes in the point (0,0).
Therefore, it only passes through II and IV quadrant.
You can also find the equation of the line, find one more point and draw it.
\(y=mx+b \\ \\
m=-32 \\
(0,0) \\
x=0 \\ y=0 \\ \\
0=-32 \times 0 + b \\
b=0 \\ \\
y=-32x\)
Find the y-coordinate of the point for example (1,y).
\((1,y) \\
x=1 \\ y \\ \\
y=-32 \times 1 \\
y=-32\)
The line passes through the points (0,0) and (1,-32). Now you can draw it (see the attachment).
The answer:
The line passes through quadrants II and IV.
