Answer:
none of them
Step-by-step explanation:
The given line is, written in a more convenient way:
[tex]y = x + 3[/tex]
I moved the (-1) to the right hand side, it became (+1) and summed with (+2) gave the (+3).
As you can see, the slope of your line is +1.
Now what do we know from the theory? We know that two lines are perpendicular if one's slope is the negative periodical of the other's.
In your case this is very easy to calculate. The negative reciprocal of +1 is -1.
The slopes of the five cases given as options are as follows:
1) -3
2) +3
3) -3
4) +1
5) -3
None of those lines is perpendicular to the original given line.
Answer:
none of them
Step-by-step explanation:
Two lines are perpendicular when satisfy the next equation: m1*m2 = -1, where m1 and m2 are the slopes o the lines.
line 1:
y – 1 = (x+2)
y = x + 3
slope of line 1 = 1
line 2:
y + 2 = –3(x – 4)
y + 2 = -3*x + 12
y = -3*x + 10
slope of line 2 = -3
m1*m2 = 1*(-3 ) = -3
They are not perpendicular
line 3:
y − 5 = 3(x + 11)
y − 5 = 3*x + 33
y = 3*x + 38
slope of line 3 = 3
m1*m3 = 1*3 = 3
They are not perpendicular
line 4:
y = -3x –
slope of line 4 = -3
m1*m4 = 1*(-3 ) = -3
They are not perpendicular
line 5:
y = x – 2
slope of line 5 = 1
m1*m5 = 1*1 = 1
They are not perpendicular
line 6:
3x + y = 7
y = -3x + 7
slope of line 6 = -3
m1*m6 = 1*(-3 ) = -3
They are not perpendicular
