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Mathematics 21 Online
OpenStudy (anonymous):

] Which equation represents the line passing through the point (4, −16) and is perpendicular to the line

OpenStudy (anonymous):

do you know the relationship between the slopes of perpendicular lines?

OpenStudy (anonymous):

the slope of your given line is -2/3. How will the slope of a line perpendicular to this line be related to -2/3?

OpenStudy (waleed_imtiaz):

I dont get ur question very well.... If u are asking about the line perpendicular to the line mentioned above y=- 2/3 x =8 , then in my view, the line perpendicular to this will have the slope= 3/2 ,

OpenStudy (anonymous):

correct. two perpendicular line having slopes m1 and m2 have slopes that are related by \[m_{1}m_{2}=-1\]

OpenStudy (anonymous):

@MeganLewis now we have the slope of our line and a point (4,-16) that it must pass through. This is all you need to get the equation of this line. Use point-slpe form.

OpenStudy (waleed_imtiaz):

very well said @eseidl now @MeganLewis Use point slope form to find the required equation of the line.......

OpenStudy (waleed_imtiaz):

@eseidl thank You for ur appreciation . Em new here....

OpenStudy (anonymous):

I'm confused..

OpenStudy (anonymous):

ok. what part? how we got the slope or you're not sure what point-slope form of a line is?

OpenStudy (anonymous):

is the answer y=3/2x-22?

OpenStudy (waleed_imtiaz):

where's the confusion....... ? and U have to apply y-y1 =m(x-x1).. and the points x1= 4 and y1=-16 and the slope is3/2.. put all the values.

OpenStudy (anonymous):

@MeganLewis looks like you made an error somewhere. Show your steps?

OpenStudy (waleed_imtiaz):

the equation of the line will be 3x-2y-44=0 .. now U can arrange it...

OpenStudy (anonymous):

maybe I made the error...let me double check.

OpenStudy (anonymous):

Yep :0 @MeganLewis your answer is correct, my bad :)

OpenStudy (anonymous):

I hope you got this answer yourself though and understand what's going on

OpenStudy (anonymous):

Thanks guys!

OpenStudy (waleed_imtiaz):

U're welcome @MeganLewis

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