– 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.

This helped A LOT. Thanks!

Great video!!

cheers

You helped Me!!

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

all you had too do was plug your problem into his problem

and work your way from there haha

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

Thank you.

How do you sole where these lines intercept?

BEST MATH CHANNEL EVER

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

Thanks for the vid

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???

thank you!

thank YOU IT LITERALLY HELP ME ALOT