A line through the points (2, -9) and (j, 17) is parallel to the line 2x + 3y = 21. What is the value of j?

j = -37

Step-by-step explanation:

First find the slope of 2x + 3y = 21

Solve for y

Subtract 2x from each side

2x-2x + 3y =-2x+ 21

3y = -2x+21

Divide by 3

3y/3 = -2x /3 + 21/3

y = -2/3 x +7

This is in slope intercept form y = mx+b where m is the slope and b is the y intercept

m = -2/3

The slope of parallel lines are equal

Using the two points

m = (y2-y1)/(x2-x1)

-2/3 = (17 - -9)/(j-2)

-2/3 = (17 +9)/(j-2)

Using cross products

-2(j-2) = 3 ( 17+9)

-2j +4 = 26*3

-2j +4 = 78

Subtract 4 each side

-2j = 78-4

-2j = 74

Divide by -2

-2j/-2 = 74/-2

j = -37

16 km due west

Step-by-step explanation:

The bearing of the school p from school q is 16 km due west.

To find the bearing of school q from school p, we have to find the direction that the school q is with respect to school p.

Since p is directly west of q, then it implies that q must be directly east of p.

We now know the direction.

Since the distance from q to p is exactly the same as the distance from p to q, then, the distance from p to q is 16 km.

Hence, the bearing of q from p is 16 km due west.