Question

Find the set of points that are part of a line that is perpendicular to the line y = ­-3x

a) (-3,-3), (3.-5)
b) (-6,-1), (3,2)
c) (-1, -1), (1, 5)
d) (3, -6), (-1, 6)

Answer 1

the answer is c. (-1, -1), (1, 5)

Explanation:

perpindicular lines have reciprocal slopes.

reciprocal terms are two identical numbers, but with opposite signs.

the slope in y= -3x is -3, and the slope in (-1, -1), (1, 5) is 3.

Answer 2

b) (-6,-1), (3,2)

Explanation:

We have an equation for a line that is:

`y=-3x`

where the number
`-3` is the slope of the line, i will call this number
`m_(1)`, so

`m_(1)=-3`

to be a perpendicular line, the slope of the new line
`m_(2)` must satisfy the following:

`m_(2)=(-1)/(m_(1))`

and since
`m_(1)=-3`

`m_(2)=(-1)/(-3)=(1)/(3)`

so now we check which one of the options is part of a line with a slope of 1/3, using the formula for the slope:

`m=(y_(2)-y_(1))/(x_(2)-x_(1))`

• a) (-3,-3), (3.-5)

`x_(1)=-3, y_(1)=-3, x_(2)=3,y_(2)=-5`

slope:
`m=(-5-(-3))/(3-(-3))=(-5+3)/(3+3)=(-2)/(6)=-(1)/(3)`, this set of points is NOT part of a perpendicular line.

• b) (-6,-1), (3,2)

`x_(1)=-6,y_(1)=-1,x_(2)=3,y_(2)=2`

slope:
`m=(2-(-1))/(3-(-6)) =(2+1)/(3+6)=(3)/(9) =(1)/(3)` this set of points are part of a perpendicular line because the slope of the line between them is
`(1)/(3)` which is the condition to be a perpendicular line.