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

Answer:

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

Step-by-step 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

Answer:

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

Step-by-step 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:

$m_{2}=\frac{-1}{m_{1}}$

and since $m_{1}=-3$

$m_{2}=\frac{-1}{-3}=\frac{1}{3}$

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:

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

• a) (-3,-3), (3.-5)

$x_{1}=-3, y_{1}=-3, x_{2}=3,y_{2}=-5$

slope: $m=\frac{-5-(-3)}{3-(-3)}=\frac{-5+3}{3+3}=\frac{-2}{6}=-\frac{1}{3}$, 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: $m=\frac{2-(-1)}{3-(-6)} =\frac{2+1}{3+6}=\frac{3}{9} =\frac{1}{3}$  this set of points are part of a perpendicular line because the slope of the line between them is $\frac{1}{3}$ which is the condition to be a perpendicular line.