Question

The two lines 3x-y-1=0 and x+3y-5=0 are:
a) perpendicular
b) parallel
c) confounded
d) collinear

Answer 1

you might wish to compare their slopes

re-writing each as \(y = ...\) would help you do that

Answer 2

i got x=3/5.
and y=4/5.
but now no idea how to continue

Answer 3

Is confounded points even a thing?

Answer 4

the right answer is a-
but how to know if 2 lines are perp/paral/conf ?

Answer 5

what you have done is found *incorrectly* the intersection point. plug your answer back into both equations to see.

as i said, if you re-write both equations in \(y = mx + c\) form, you will see m, ie the slope, for each. we can go from there.

[and i know that n some countries they use \(y = ax + b\) or \(y = mx + b\) but do that first....]

Answer 6

@Sepeario i don't think so. i actually looked it up :p

Answer 7

oki so i got the first y1=3x-1
and y2=(-x/3)-(5/3)
what is the next step?

Answer 8

y2 has an error in it

but this is the missing bit of information you need

when you multiply the slopes of lines that are perpendicular, you get -1

https://gyazo.com/1fa13ef12ef26796eab06a832615537a

Answer 9

ohhh thank you so much you were so helpful.
A little more question,
how to know if they are parallel?

Answer 10

FYI, you should get the equations
\[ y = 3x -1 \\y= -\frac{1}{3} x + \frac{5}{3} \]
compare the slopes: if they are the same the lines are parallel (unless you have the exact same equation for both, in which case they are the same line)
if the slopes are *negative reciprocals*, the lines are perpendicular (form a right angle where they cross)

negative reciprocal means "flip" and multiply by -1

for examples: 2/3 and -3/2 (first flip 2/3 to get 3/2 , then negate to get -3/2)
or 2 and -1/2 : flip 2 (think of it as 2/1) to get 1/2 , then negate: -1/2

or -1/4 and 4
etc

the other test is if you multiply them you get -1 as Irish explained up above