Ex 2: Find the Equation of a Line Perpendicular to a Given Line Passing Through a Given Point

Ex 2:  Find the Equation of a Line Perpendicular to a Given Line Passing Through a Given Point


– WE’RE GOING TO DETERMINE
THE EQUATION OF THE LINE THAT IS PERPENDICULAR
TO THE LINE GIVEN BY THE EQUATION X – 4Y=15 AND PASSES THROUGH
THE POINT 3, 1. WE WANT THE LINEAR EQUATION
IN SLOPE INTERCEPT FORM. SO JUST FOR A QUICK REVIEW,
IF TWO LINES ARE PERPENDICULAR THEY INTERSECT AND FORM A RIGHT
ANGLE OR A 90 DEGREE ANGLE, AND THEIR SLOPES ARE NEGATIVE
RECIPROCALS OF ONE ANOTHER WHICH CAN BE WRITTEN
USING THIS NOTATION HERE WHICH CAN BE A LITTLE CONFUSING. SO FOR EXAMPLE IF THIS
FIRST LINE HAS A SLOPE OF 2/3, THEN THE LINE PERPENDICULAR
TO THIS LINE WOULD HAVE A SLOPE OF -3/2. SO TO DETERMINE
A NEGATIVE RECIPROCAL, NOTICE HOW WE FLIP THE FRACTION
OVER AND CHANGE THE SIGN. SO OUR FIRST STEP
IS GOING TO BE TO DETERMINE THE SLOPE OF THE GIVEN LINE BY WRITING THIS EQUATION
IN SLOPE INTERCEPT FORM WHERE THE FORM Y=MX + B WHERE M IS THE SLOPE
OF THE LINE. SO IF WE HAVE X – 4Y=15, WE NEED TO SOLVE THIS EQUATION
FOR Y. SO WE’LL START BY SUBTRACTING X
ON BOTH SIDES. SO WE’LL HAVE -4Y EQUALS–
WE’LL WRITE THIS AS -X + 15, AND THEN THE LAST STEP, WE’LL DIVIDE BOTH SIDES
OF THE EQUATION BY -4. SO WE’LL HAVE Y EQUALS–NOW
TO SIMPLIFY THIS, WE COULD WRITE -X AS -1X
AND -1 DIVIDED BY -4=1/4. SO WE’LL HAVE 1/4X – 15/4. SO NOTICE THE SLOPE OF THIS
LINE, WE’LL CALL IT “SLOPE ONE,” IT’S GOING TO BE EQUAL TO 1/4 WHICH MEANS IF WE WANT A LINE
THIS PERPENDICULAR TO THIS, WE NEED THE SLOPE THAT’S
THE NEGATIVE RECIPROCAL OF THIS. SO IF THEY FLIP THIS OVER
AND CHANGE THE SIGN, WE WOULD HAVE A SLOPE OF -4/1
WHICH SIMPLIFIES TO -4. SO NOW IT COMES DOWN
TO DETERMINING THE EQUATION OF A LINE THAT HAS A SLOPE OF -4 AND
PASSES THROUGH THE POINT 3, 1. AND WE’LL DO THIS USING
THE POINT SLOPE FORM OF ALIGN WHICH IS Y – Y SUB 1
=M x THE QUANTITY X – X SUB 1 WHERE M IS THE SLOPE OF THE LINE
AND X SUB 1, Y SUB 1 ARE THE COORDINATES OF ANY POINT
ON THAT LINE. SO FOR OUR EXAMPLE, X SUB 1=3
AND Y SUB 1=1. SO WE’LL HAVE Y – 1 MUST EQUAL
-4 x THE QUANTITY X – 3. NOW THIS IS THE EQUATION
OF THE LINE THAT WE NEED BUT SINCE WE DO WANT THIS
IN SLOPE INTERCEPT FORM, WE NOW NEED TO SOLVE THIS FOR Y. SO TO CLEAR THE PARENTHESIS
WE’LL DISTRIBUTE SO WE’LL HAVE Y – 1 EQUALS–
THIS WILL BE -4X + 12 AND THEN WE’LL ADD 1
TO BOTH SIDES TO SOLVE FOR Y. SO THE EQUATION OF OUR LINE IS
PERPENDICULAR TO THE GIVEN LINE PASSING THROUGH THE POINT 3, 1
WOULD BE Y=-4X + 13. WELL THAT WAS QUITE A BIT
OF WORK. LET’S GO AHEAD AND VERIFY
THIS GRAPHICALLY. LET’S MAKE SURE THAT THIS LINE IS PERPENDICULAR
TO THIS LINE HERE AND OUR LINE PASSES
THROUGH THE POINT 3, 1. SO THE RED LINE IS A GRAPH
OF THE GIVEN LINE. REMEMBER IT HAD A SLOPE OF 1/4
AND Y INTERCEPT OF -15/4 AND THE EQUATION THAT WE FOUND
WAS Y=-4X + 13. NOTICE HOW THESE TWO LINES DO
INTERSECT AND FORM A RIGHT ANGLE AND WE KNOW THAT TO BE TRUE
BECAUSE THESE SLOPES ARE NEGATIVE RECIPROCALS
OF ONE ANOTHER AND MORE IMPORTANTLY
NOTICE HOW THE GRAPH OF OUR LINE DOES CONTAIN THE POINT 3, 1
WHICH WAS GIVEN IN THE PROBLEM. SO THIS DOES VERIFY
THAT OUR WORK IS CORRECT.

15 thoughts on “Ex 2: Find the Equation of a Line Perpendicular to a Given Line Passing Through a Given Point

  • September 1, 2012 at 11:02 pm
    Permalink

    This helped A LOT. Thanks!

    Reply
  • November 20, 2012 at 1:06 pm
    Permalink

    Great video!!

    Reply
  • November 28, 2012 at 9:54 pm
    Permalink

    cheers

    Reply
  • January 16, 2013 at 9:35 pm
    Permalink

    You helped Me!!

    Reply
  • February 5, 2013 at 5:03 pm
    Permalink

    Through (2, -8) and perpendicular to the line 2x-3y=5. how about this? can you do it?

    Reply
  • December 2, 2013 at 2:57 am
    Permalink

    all you had too do was plug your problem into his problem
    and work your way from there haha

    Reply
  • December 17, 2013 at 3:46 pm
    Permalink

    you should be my math teacher, this helped a lot thanks!!

    Reply
  • November 24, 2014 at 5:08 am
    Permalink

    Thank you.

    Reply
  • September 17, 2015 at 1:49 pm
    Permalink

    How do you sole where these lines intercept?

    Reply
  • October 13, 2015 at 11:08 pm
    Permalink

    BEST MATH CHANNEL EVER

    Reply
  • May 3, 2017 at 5:17 am
    Permalink

    IGCSE Math Exam in ONE HOUR!!! … Any One Else

    Reply
  • September 5, 2017 at 8:32 pm
    Permalink

    Thanks for the vid

    Reply
  • March 3, 2018 at 12:18 am
    Permalink

    thanks for the video, just wanted to ask you why did you flip the fraction of m1 in order to get m2 is that a rule we have to follow to get m2???

    Reply
  • April 21, 2018 at 2:58 am
    Permalink

    thank you!

    Reply
  • September 22, 2018 at 2:32 am
    Permalink

    thank YOU IT LITERALLY HELP ME ALOT

    Reply

Leave a Reply

Your email address will not be published. Required fields are marked *