Ex 1: Find the Equation of a Line Perpendicular to a Given Line Passing Through a Given Point

Ex 1:  Find the Equation of a Line Perpendicular to a Given Line Passing Through a Given Point


– WE WANT TO DETERMINE
THE EQUATION OF A LINE THAT IS PERPENDICULAR
TO Y=2X + 1, AND PASSES THROUGH THE POINT
WITH COORDINATES (-4,5). IF TWO LINES ARE PERPENDICULAR, THAT MEANS THEY INTERSECT
AND FORM A RIGHT ANGLE OR A 90 DEGREE ANGLE. AND IF TWO LINES
ARE PERPENDICULAR, THEN THEIR SLOPES ARE NEGATIVE
RECIPROCALS OF ONE ANOTHER, WHICH CAN BE WRITTEN
USING THIS NOTATION HERE, WHICH CAN BE
A LITTLE BIT CONFUSING. SO FOR EXAMPLE,
IF THIS FIRST LINE HAD A SLOPE OF LET’S SAY, 2/5, THEN THE SLOPE
OF THE SECOND LINE HERE WOULD HAVE TO BE
THE NEGATIVE RECIPROCAL, MEANING WE’LL FLIP THIS OVER
AND THEN CHANGE THE SIGN. SO IT’S GOING TO BE -5/2. OKAY, SO BACK TO OUR PROBLEM, WE WANT A LINE THAT IS
PERPENDICULAR TO Y=2X + 1. WELL, THIS LINE
IS IN SLOPE INTERCEPT FORM, SO WE SHOULD RECOGNIZE THAT
THE SLOPE OF THIS LINE IS 2/1. THEREFORE, THE SLOPE
OF THE PERPENDICULAR LINE WOULD BE THE NEGATIVE RECIPROCAL
OF THIS. SO FOR THE SLOPE OF OUR LINE, WE’LL HAVE TO FLIP THIS OVER
AND CHANGE THE SIGN. THAT WOULD GIVE US -1/2. -1/2 IS A NEGATIVE RECIPROCAL
OF 2/1. AND THEN WE KNOW OUR LINE ALSO
PASSES THROUGH THE POINT (-4,5). NOW, TO DETERMINE THE EQUATION
OF THE LINE WITH SLOPE OF -1/2 CONTAINING THE POINT (-4,5), WE CAN EITHER USE
SLOPE INTERCEPT FORM, OR THE POINT SLOPE FORM
OF THE LINE. AND I THINK FOR THIS VIDEO
WE’LL SHOW BOTH. SO USING JUST SLOPE INTERCEPT
FORM OR THE FORM Y=MX + B, WE WOULD FIRST SUBSTITUTE -1/2
FOR M. SO WE’D HAVE Y=-1/2X + B. AND THEN TO DETERMINE
THE Y INTERCEPT, IT’S NOT GOING TO BE +5 BECAUSE
THE X COORDINATE IS NOT ZERO. BUT SINCE THIS POINT
IS ON THE LINE THAT MEANS IF WE SUBSTITUTE
THESE COORDINATES INTO THE EQUATION IT SHOULD SATISFY THE EQUATION. SO WE’LL SUBSTITUTE -4 FOR X,
5 FOR Y, AND THEN SOLVE FOR B. SO WE’D HAVE 5=-1/2 x -4 + B. PUT THAT -4/1
SO WE HAVE 5 EQUALS– AND SO THIS SIMPLIFIES HERE,
THERE’S ONE 2 IN 2, AND TWO 2s IN 4. THIS BECOMES -1 x -2. DON’T FORGET
ABOUT THE SIGN HERE. SO NOW IF WE SUBTRACT 2
ON BOTH SIDES– WE HAVE B EQUALS–
THIS WILL BE +3. SO NOW WE HAVE OUR EQUATION, AND
WE’LL JUST SUBSTITUTE +3 FOR B. AND THE EQUATION WILL BE
Y=-1/2X + 3. AND LET’S GO AHEAD AND SEE
IF WE CAN GET THE SAME EQUATION USING POINT SLOPE FORM. POINT SLOPE FORM OF A LINE IS
Y – Y1=M(X – X SUB 1), WHERE M IS THE SLOPE, AND (X SUB 1,Y SUB 1)
IS A POINT ON THE LINE. SO FOR THIS SITUATION
X SUB 1 IS GOING TO BE -4, AND Y SUB 1 WILL BE +5. SO WE’D HAVE Y – 5=-1/2(X). NOW, HERE WE’D HAVE MINUS -4,
WHICH IS THE SAME AS +4. AND NOW WE’LL GO AHEAD
AND SOLVE THIS EQUATION FOR Y. SO WE’LL DISTRIBUTE HERE, SO WE’LL HAVE Y – 5=
THIS WOULD BE -1/2X. THEN HERE WE’D HAVE -2, AND THE LAST STEP TO SOLVE FOR Y
WOULD BE TO ADD 5. THIS WOULD BE 0,
SO WE’D HAVE Y=-1/2X, AND THIS WILL BE +3. SO NOTICE HOW IF WE JUST USE
SLOPE INTERCEPT FORM TO DETERMINE THE EQUATION, OR IF WE USE POINT SLOPE FORM
TO DETERMINE THE EQUATION, THE RESULT IS THE SAME. AND LET’S GO AHEAD AND QUICKLY
VERIFY THIS GRAPHICALLY. YOU WERE GIVEN THE RED LINE
Y=2X + 1, AND WE DETERMINED
THAT THE EQUATION OF THE PERPENDICULAR LINE
WAS Y=-1/2X + 3. NOTICE HOW THESE TWO LINES DO
INTERSECT IN A RIGHT ANGLE HERE, AND THE BLUE LINE DOES CONTAIN
THE GIVEN POINT THAT HAD A COORDINATES (-4,5),
THIS POINT HERE. SO OUR WORK LOOKS GOOD. I HOPE YOU FOUND THIS HELPFUL.

99 thoughts on “Ex 1: Find the Equation of a Line Perpendicular to a Given Line Passing Through a Given Point

  • November 13, 2013 at 12:36 am
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    Thank you soo much! You literally saved my life! My gratitude for posting this video <3

    Reply
  • November 24, 2013 at 8:01 pm
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    Thank you

    Reply
  • November 25, 2013 at 1:56 am
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    Thank you, Yes I can finish my work!

    Reply
  • December 2, 2013 at 10:07 pm
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    Perfect for tomorrows test!

    Reply
  • December 11, 2013 at 3:18 am
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    Thank u so much

    Reply
  • December 17, 2013 at 11:45 pm
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    I need some engineering analytic geometry sir πŸ™‚ pls upload some

    Reply
  • February 2, 2014 at 8:02 pm
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    Thanks

    Reply
  • February 12, 2014 at 11:38 pm
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    Thank You, you saved my math grade!

    Reply
  • May 5, 2014 at 1:04 am
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    thanks

    Reply
  • May 22, 2014 at 12:40 am
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    Holy shiz THANKS…

    Reply
  • June 3, 2014 at 9:03 pm
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    Thank you so much! Need this for the NYS Regents!

    Reply
  • June 6, 2014 at 2:56 pm
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    THANK YOU got maths gcse monday :((((

    Reply
  • June 23, 2014 at 7:22 pm
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    This was so helpful. Thank you~~

    Reply
  • September 4, 2014 at 3:26 pm
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    OMG THANKS A BUNCH MATE !!!!!!!!! Β Love you, no homo <3

    Reply
  • September 27, 2014 at 1:48 pm
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    Thank you so very much!

    Reply
  • October 19, 2014 at 3:42 pm
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    THANK YOU SO MUCH!!!!

    Reply
  • December 10, 2014 at 10:40 am
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    Do u not need to do anything with the one

    Reply
  • December 11, 2014 at 9:12 pm
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    OMFG THANK YOU SOOOOOOOOO MUCH!!!!!!!!!!!!!!!!!!!

    Reply
  • March 25, 2015 at 9:52 pm
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    Thanks! Great help.

    Reply
  • April 19, 2015 at 11:01 pm
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    Thanks! You helped me so much… You are a really good explainer.

    Reply
  • May 18, 2015 at 6:05 pm
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    how did you get 3 on the first method when the difference between 5 and -2 is 8

    Reply
  • June 7, 2015 at 11:37 pm
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    My GCSE exam is in less than 9 hours. Thank you so much.

    Reply
  • June 25, 2015 at 8:02 pm
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    Well, since everyone is saying thanks….
    THANKS! I'm taking Algebra II next year and i don't remember this AT ALL!
    Thanks very much!

    Reply
  • October 7, 2015 at 2:17 am
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    Great video!

    Reply
  • October 12, 2015 at 2:26 pm
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    wish u were my math professor..

    Reply
  • October 21, 2015 at 2:20 am
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    Thank you very much. It was really very helpful

    Reply
  • October 26, 2015 at 3:31 am
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    Doing a review sheet currently and thus really helped me out thanks!

    Reply
  • November 4, 2015 at 5:28 am
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    Great video, look forward to seeing more! You helped me understand the concept a lot better, thanks for the help b4 my test!

    Reply
  • November 8, 2015 at 7:57 pm
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    thank you dear

    Reply
  • November 16, 2015 at 1:33 am
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    I like how most of the people using this video are in highschool of college, I'm in 7th grade advanced algebraπŸ˜‚πŸ˜‚

    Reply
  • December 2, 2015 at 2:56 am
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    What would u do if they only gave u x and the points

    Reply
  • December 2, 2015 at 4:28 am
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    Thanks very helpful.

    Reply
  • December 3, 2015 at 4:31 am
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    thank you for this my final is tomorrow and i was so stuck so helpful

    Reply
  • December 4, 2015 at 2:56 am
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    Thanks so much this helped me a lot

    Reply
  • December 9, 2015 at 9:36 pm
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    Yo yo good lookin out playa

    Reply
  • December 9, 2015 at 11:27 pm
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    great video and big help! thank you

    Reply
  • December 13, 2015 at 12:08 am
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    thanks helped out a lot

    Reply
  • January 14, 2016 at 12:39 am
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    This was really helpful thanks so much

    Reply
  • January 28, 2016 at 4:21 am
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    Thank you so much for this video! The clear and concise explanations were extremely helpful.

    Reply
  • February 18, 2016 at 6:12 pm
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    Thank you!

    Reply
  • March 2, 2016 at 6:18 pm
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    Brilliant video!

    Reply
  • March 3, 2016 at 5:11 pm
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    Thank you for captioning this

    Reply
  • April 17, 2016 at 6:21 am
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    we appreciate you

    Reply
  • May 10, 2016 at 3:41 am
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    Thankyouuu. Its very helpful.

    Reply
  • May 15, 2016 at 10:29 pm
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    This was helpful.

    Reply
  • May 20, 2016 at 9:48 pm
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    great video, very helpful.

    Reply
  • May 22, 2016 at 4:48 pm
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    but what if the two fractions cannot simplify?

    Reply
  • July 3, 2016 at 7:48 am
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    Very helpful, thanks πŸ™‚

    Reply
  • August 3, 2016 at 4:19 pm
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    thanks a lot 😍😍

    Reply
  • September 19, 2016 at 8:33 pm
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    This was helpful

    Reply
  • September 28, 2016 at 8:04 pm
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    Wow great video man! Thank you!!

    Reply
  • October 27, 2016 at 2:43 am
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    to boring to watch ! had to repeat this video 6 times lol

    Reply
  • October 27, 2016 at 3:08 am
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    You've raised my grade from a 60 to a 90!! thanks alot!!! I understand this!

    Reply
  • December 21, 2016 at 2:07 am
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    Awesome video, loved it! Thank you, this really helped me study for tomorrows test

    Reply
  • December 25, 2016 at 1:13 am
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    Thank you that video it's very greatful…..yeah…

    Reply
  • January 18, 2017 at 11:47 pm
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    3:00 What About The +1???

    Reply
  • February 23, 2017 at 4:51 am
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    Thank you, such a life saver 😊

    Reply
  • May 2, 2017 at 5:32 pm
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    Bro thanks

    Reply
  • May 15, 2017 at 8:21 am
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    if only i saw this a day earlier
    .. i messed up a test at school, and now i am depressed..

    Reply
  • May 20, 2017 at 4:35 pm
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    You sir are my god and saviour

    Reply
  • May 26, 2017 at 8:57 am
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    Thanks! I have an exam soon and I'm so confused about most of the Linear section.

    Reply
  • August 30, 2017 at 12:40 am
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    Very neat + easy explanation!! Thanks a lot for posting this πŸ™‚

    Reply
  • August 31, 2017 at 1:18 pm
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    Thanks , good method helped me out!

    Reply
  • September 18, 2017 at 10:38 pm
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    You da real MVP

    Reply
  • October 31, 2017 at 2:16 am
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    thank you so much

    Reply
  • November 27, 2017 at 9:03 pm
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    Please send me emails of the videos. Im stuck at school.

    Reply
  • December 7, 2017 at 1:16 am
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    thank you so much I was overthinking this way too much

    Reply
  • December 13, 2017 at 12:23 am
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    thx

    Reply
  • January 7, 2018 at 10:58 pm
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    Thnks

    Reply
  • January 8, 2018 at 4:10 am
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    THANK YOU SO MUCH AND HAPPY NEW YEAR 2018!!!

    Reply
  • February 1, 2018 at 2:08 am
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    Thx so much

    Reply
  • February 9, 2018 at 3:01 am
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    thanks man !

    Reply
  • April 12, 2018 at 2:20 am
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    thank you I was struggling on my homework and now I got it thank you again !

    Reply
  • April 19, 2018 at 4:40 pm
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    very helpful

    Reply
  • May 6, 2018 at 8:04 pm
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    Thx a ton bro, maths can get so hard sometimes

    Reply
  • May 29, 2018 at 4:10 am
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    This is soooo helpful for my upcoming math unit 6 secondary 1 test tomorrow! All the visuals really make more sense all laid out.

    Reply
  • June 7, 2018 at 9:03 am
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    There's just one thing I dont get, isnt this the same as finding a parellel line equation?

    Reply
  • August 24, 2018 at 5:05 am
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    yes!!

    Reply
  • August 29, 2018 at 5:21 am
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    what if its -8x will the reciprocal be -1/8x or 1/8x? someone help PLEASE!! have to prepare for a test

    Reply
  • September 5, 2018 at 8:29 am
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    its awesome

    Reply
  • September 6, 2018 at 11:52 pm
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    Super helpful. Thanks!!!

    Reply
  • October 8, 2018 at 6:37 pm
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    Not enough explanation lol

    Reply
  • October 13, 2018 at 4:06 pm
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    I'm taking algebra this year and I'm about to cry in class because the teacher says a bunch of useless information like foods and his family……
    THANK YOU SOOOOOOOOO MUCH for this video!

    Reply
  • October 29, 2018 at 3:16 am
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    Thank you

    Reply
  • November 11, 2018 at 2:31 am
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    Thanks

    Reply
  • November 17, 2018 at 11:57 am
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    thank you so much .

    Reply
  • November 28, 2018 at 4:09 am
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    Help!

    Write an equation in slope intercept form for the line that
    passes through the given point and is perpendicular to the graph of the
    equation.

    (-2,3),y=-1/2x-4

    Reply
  • January 27, 2019 at 2:07 am
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    thx bro that helped a lot for midterms!!

    Reply
  • February 20, 2019 at 6:31 am
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    Thanks for this video.

    Reply
  • August 11, 2019 at 11:19 pm
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    I got an a+++++-+β€”+ from this thanks πŸ™

    Reply
  • September 13, 2019 at 2:17 am
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    what if the slope is 1? would you just use -1?

    Reply
  • October 3, 2019 at 3:32 am
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    Thank you so much

    Reply
  • November 19, 2019 at 2:32 am
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    Once again, YouTube has taught be better than my own teacher!

    Reply
  • November 22, 2019 at 12:24 pm
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    Thanks πŸ™

    Reply
  • November 22, 2019 at 5:23 pm
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    the internet is going to be the new school. prediciton 2030

    Reply
  • November 27, 2019 at 3:07 pm
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    thank you, ill use this for programming algorithm

    Reply
  • December 18, 2019 at 4:49 pm
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    THANK YOU SO MUCH YOU SAVED ME

    Reply

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