- WELCOME TO OUR SECOND VIDEO
ON PARALLEL AND PERPENDICULAR LINES.
IN THIS VIDEO, WE'LL DO A QUICK REVIEW OF THE DEFINITIONS
OF PARALLEL AND PERPENDICULAR LINES.
AND THEN WE'LL FOCUS ON DETERMINING THE EQUATIONS
OF PARALLEL AND PERPENDICULAR LINES
BASED UPON GIVEN INFORMATION.
PARALLEL LINES ARE TWO LINES THAT NEVER INTERSECT.
AND IN ORDER FOR THAT TO HAPPEN,
THE TWO LINES MUST HAVE THE SAME SLOPE AND DIFFERENT INTERCEPTS.
SO THE SLOPES MUST BE THE SAME,
AND THE Y-INTERCEPTS MUST BE DIFFERENT.
AND HERE'S AN ILLUSTRATION OF TWO PARALLEL LINES.
USING THESE TWO POINTS, WE CAN DETERMINE THE SLOPE OF THIS LINE
WOULD BE UP ONE, RIGHT TWO, SO 1/2.
AND USING THESE TWO POINTS ON THE RED LINE,
WE COULD ALSO DETERMINE THE SLOPE IS EQUAL TO 1/2.
AND WE CAN ALSO TELL THAT THE Y-INTERCEPTS ARE DIFFERENT,
SO WE HAVE PARALLEL LINES.
PERPENDICULAR LINES ARE TWO LINES THAT INTERSECT
AND FORM A RIGHT ANGLE.
IN ORDER FOR TWO LINES TO BE PERPENDICULAR,
THE PRODUCT OF THE SLOPES MUST EQUAL -1,
OR WE CAN ALSO SAY THAT THE SLOPES ARE NEGATIVE RECIPROCALS.
SO, FOR EXAMPLE, IF THE SLOPE OF LINE ONE IS EQUAL TO 3/4,
THEN THE SLOPE OF THE SECOND LINE
THAT WOULD BE PERPENDICULAR TO THIS LINE WOULD BE -4/3.
SO WE FLIP IT OVER AND CHANGE THE SIGN.
AND HERE'S AN ILLUSTRATION OF TWO PERPENDICULAR LINES.
THEY INTERSECT AND FORM A RIGHT ANGLE,
WHICH IS EQUAL TO 90 DEGREES.
AND WE CAN SEE THAT THE SLOPE OF THE BLUE LINE
WOULD BE EQUAL TO 1/2 USING THESE TWO POINTS.
AND USING THESE TWO POINTS,
WE CAN DETERMINE THE SLOPE OF THE RED LINE
WOULD BE DOWN TWO, RIGHT ONE, WHICH WOULD BE -2/1.
SO WE CAN SEE THEY'RE NEGATIVE RECIPROCALS,
WHICH OF COURSE DOES EQUAL -2. OKAY.
BUT THE FOCUS OF THIS VIDEO IS TO DETERMINE EQUATIONS OF LINES
BASED UPON GIVEN INFORMATION.
SO ON THIS QUESTION, WE'RE GOING TO DETERMINE THE EQUATION
OF A LINE THAT IS PARALLEL TO THIS LINE
AND PASSES THROUGH THE POINT (6, 2).
SO THE FIRST THING WE'RE GOING TO DO IS TAKE THE GIVEN LINE
AND DETERMINE WHAT THE SLOPE IS.
AND WE'RE GOING TO DO THAT BY SOLVING THIS EQUATION FOR Y.
SO WE'LL START BY SUBTRACTING X ON BOTH SIDES.
IT'S OFTEN VERY CONVENIENT TO HAVE A COEFFICIENT FOR X,
SO WE'RE GOING TO GO AHEAD AND PUT A ONE HERE.
AND THEN WE'LL DIVIDE EVERYTHING BY -3.
SO WE HAVE Y EQUALS-- THIS WOULD BE +1/3X - 3,
SO THIS TELLS US THE SLOPE OF THE GIVEN LINE IS EQUAL TO 1/3.
NOW, SINCE THE LINE THAT WE WANT IS PARALLEL TO THIS LINE,
WE KNOW THE SLOPE OF THE NEW LINE IS ALSO GOING TO BE 1/3.
SO NOW WE WANT TO FIND THE EQUATION OF THE LINE
WHERE THE SLOPE IS EQUAL TO 1/3
AND IT PASSES THROUGH THE POINT (6, 2).
NOW WE CAN DO THIS EITHER USING SLOPE-INTERCEPT FORM
OR POINT-SLOPE FORM.
LET'S START BY USING SLOPE-INTERCEPT FORM.
I MEAN, WE'RE GOING TO FIND THE EQUATION OF THE LINE
IN THIS FORM,
AND WE KNOW THAT M IS EQUAL TO 1/3.
SO WE HAVE Y = 1/3X + B.
NOW, SINCE THIS POINT IS ON THE LINE,
THAT MEANS IT WOULD BE A SOLUTION TO THIS EQUATION.
SO WE CAN REPLACE X WITH SIX, Y WITH TWO AND THEN SOLVE FOR B.
SO USING THE POINT (6, 2), WE WOULD HAVE 2 = 1/3 x 6 + B.
SO WE HAVE 2 = 1/3 x 6/1. THAT WOULD BE TWO.
SO IF WE SUBTRACT TWO ON BOTH SIDES,
IT LOOKS LIKE B WOULD BE EQUAL TO ZERO,
WHICH TELLS US THE EQUATION OF OUR LINE
THAT IS PARALLEL TO THE GIVEN LINE
AND PASSES THROUGH THE POINT (6, 2)
WOULD BE Y = 1/3X + 0 OR JUST Y = 1/3X.
NOW, I DO ALSO WANT TO SHOW THIS USING POINT-SLOPE FORM.
SO INSTEAD OF USING Y = MX + B,
WE COULD HAVE USED Y - Y1 = M x (X - X1).
THIS IS--AGAIN, THIS IS POINT-SLOPE FORM.
SO USING THIS FORM, WE'D HAVE Y - Y1, WHICH IS 2,
WOULD EQUAL 1/3 x (X - X1), WHICH IS EQUAL TO SIX.
AND NOW WE HAVE TO DISTRIBUTE THE 1/3
AND THEN ADD TWO TO BOTH SIDES.
SO WE'D HAVE 1/3X. 1/3 x -6 WOULD BE -2, SO -2.
AND NOW WE'LL ADD TWO TO BOTH SIDES.
AND AS WE EXPECTED, THE RESULT IS THE SAME.
NOW LET'S GO AHEAD AND VERIFY THIS ON THE GRAPHING CALCULATOR.
WE'RE GOING TO ENTER IN THE GIVEN LINE IN Y1,
AND WE'LL COMPARE IT Y = 1/3X.
WE PRESS Y EQUALS--WE'RE GOING TO TYPE IN PARENTHESES 1/3X - 3.
THEN Y2 WILL BE 1/3X.
AND WE'LL GO AHEAD AND PRESS GRAPH.
THE SLOPES ARE THE SAME,
AND THEREFORE WE HAVE PARALLEL LINES.
NOW WE WANT TO DETERMINE THE EQUATION OF A LINE
THAT IS PERPENDICULAR TO THIS LINE
AND PASSES THROUGH THE POINT (3, 4).
SO, AGAIN, THE FIRST STEP IS TO DETERMINE THE SLOPE
OF THE GIVEN LINE.
FROM THAT WE CAN DETERMINE THE SLOPE OF A LINE
THAT WILL BE PERPENDICULAR TO THAT.
SO WE'LL SUBTRACT 2X ON BOTH SIDES.
AND 3Y = -2X - 6.
WE'LL DIVIDE EVERYTHING BY THREE.
SO THE GIVEN LINE WOULD BE Y = - 2/3X - 2.
SO THIS SLOPE IS EQUAL TO -2/3.
WELL, THE LINE THAT WE'RE LOOKING FOR
IS PERPENDICULAR TO THIS,
SO THAT MEANS THE SLOPE THAT WE WANT TO USE
WOULD BE THE NEGATIVE RECIPROCAL OR 3/2.
SO WE WANT TO FIND THE EQUATION OF THE LINE THAT HAS SLOPE 3/2
AND PASSES THROUGH THE POINT (3, 4).
AGAIN, WE CAN USE POINT-SLOPE FORM,
OR WE CAN USE SLOPE-INTERCEPT FORM.
I'M GOING TO GO AHEAD AND USE SLOPE-INTERCEPT FORM.
SO WE KNOW THE SLOPE IS 3/2.
SO WE KNOW THAT Y IS EQUAL TO 3/2X + B.
THIS POINT IS ON THE LINE,
THEREFORE THIS POINT IS A SOLUTION
TO THIS LINEAR EQUATION.
SO WE'LL REPLACE Y WITH FOUR AND X WITH THREE.
I'M GOING TO PUT 3/1 SINCE IT'S A FRACTION,
SO WE HAVE FOUR EQUALS-- THIS WOULD BE 9/2 + B.
SO WE'LL SUBTRACT 9/2 ON BOTH SIDES.
SO WE HAVE B EQUALS--
WELL, FOUR IS ACTUALLY 8/2.
8/2 - 9/2 WOULD BE -1/2.
SO THAT TELLS US THAT THE EQUATION OF THE LINE
THAT'S PERPENDICULAR TO THE GIVEN LINE
AND PASSING THROUGH (3, 4)
IS Y = 3/2X - 1/2.
LET'S TAKE A MOMENT TO VERIFY THAT AS WELL GRAPHICALLY.
SO ENTER THESE TWO EQUATIONS INTO THE GRAPHING CALCULATOR,
CLEAR THESE OUT.
FIRST LINE HAS A SLOPE OF -2/3.
SECOND LINE WAS 3/2X - 1/2.
REMEMBER TO CHECK FOR PERPENDICULAR LINES.
WE DO HAVE TO SQUARE THE WINDOW BY PRESSING ZOOM, OPTION FIVE,
FOR Z SQUARE.
OTHERWISE, THE ANGLE BETWEEN THE TWO LINES WILL BE DISTORTED.
AND WE CAN SEE HERE THOSE LOOK LIKE THEY ARE PERPENDICULAR.
LET'S TAKE A LOOK AT ONE MORE.
DETERMINE THE EQUATION OF A LINE THAT IS PERPENDICULAR TO Y = 4
AND PASSES THROUGH THE POINT (-3, -2).
Y = 4 IS A HORIZONTAL LINE THAT LOOKS LIKE THIS.
WELL, THE SLOPE OF A HORIZONTAL LINE IS EQUAL TO ZERO,
AND THE RECIPROCAL OF ZERO IS UNDEFINED.
SO LET'S THINK ABOUT THIS A LITTLE BIT MORE.
THE ONLY THAT'S PERPENDICULAR TO HORIZONTAL LINE
WOULD BE A VERTICAL LINE.
SO LET'S PLOT THIS POINT HERE
AND FIGURE OUT WHAT THE EQUATION WOULD NEED TO BE.
HERE'S THE POINT (-3, -2).
SO THE ONLY LINE THAT WOULD PASS THROUGH THIS POINT
PERPENDICULAR TO THIS HORIZONTAL LINE
WOULD BE THIS VERTICAL LINE HERE.
SO THE QUESTION BECOMES,
WHAT IS THE EQUATION OF THIS VERTICAL LINE?
REMEMBER THAT THE EQUATION OF ANY VERTICAL LINE
IS IN THE FORM OF X EQUALS A CONSTANT.
SO IF IT PASSES THROUGH THE POINT (-3, -2),
THE EQUATION MUST BE X = -3.
OKAY. THAT'S GOING TO DO IT FOR PART TWO.
I HOPE YOU FOUND THIS VIDEO HELPFUL.