math

we know that y-y1=m(x-x1)

Assuming (2, 5) = (x1, y1) and m = 3 and substituting the values, we get: y-5=3(x-2)

3x-y-1=0 ( this is the equation )

Part B : replace x by 7

3*7-1=y 

y=20

so the path doesn't pass through the point (7;22)

but (7:20)

Part C 

the second equation can be written as y=(-1/3) x + 2

so the multiple of their m is -1

so the equations are perpendicular  to each other

f a ship's path is mapped on a coordinate grid, it follows a straight-line path of slope 3 and passes through point (2, 5).

Part A: What is the equation of the path?

Part B: Does the ship pass through point (7, 22)?

Part C: A second ship follows a straight line, with the equation x + 3y − 6 = 0. Are these two ships sailing perpendicular to each other?

The equation of the first line is given by :

y - 5 = 3(x - 2)     simplify

y = 3x - 6 + 5

y = 3x - 1

For part B, substitute  7 for x in the equation and we have :  3(7) - 1  =  21 - 1  = 20......so......the ship passes through (7, 20), but not (7,22)

Part C......x + 3y - 6  = 0   .....rearrange this as      y = (-1/3)x + 2..........the slope of this line is the negative reciprocal of the first line, so they are perpendicular.......