G 9 M 12 L 8 1 Line Relationship using Slope

G 9 M 12 L 8 1 Line Relationship using Slope

النص الكامل للفيديو

the eighth listening module 12 is about slope and equations of lines the learning objectives of this lesson are classify lines as parallel perpendicular or neither by comparing the slopes of the lines and then to classify lines as parallel perpendicular neither by comparing the equations of the lines just by the slopes or by the equations you can tell whether the lines are parallel or perpendicular or neither so we will see how to do them before that keywords of this lesson are parallel perpendicular slope we have come across this before already now we will go through the Mickey slope criteria for parallel and perpendicular lines now slope is the ratio of the change in Rise that is coordinate that is rise is always vertical to the corresponding change in the Run means horizontal so DX by is slope now slope criteria suggests method for proving the relationship between lines based on the comparison of slopes of the lines so now over here this is very important let us see for example based on the slope criteria if there are two lines how are they parallel or avadi perpendicular if the slope is same if the parallel lines have same slope then it is parallel if the two lines are same slope then it is parallel Line This is slope criteria basically based on the slope or the relationship of the slope we can tell what lines are they same way for perpendicular we can tell perpendicular slopes are based on negative reciprocal or imagine you have the slopes of these two lines now just give example 1 by 3 and -3 whenever you multiply these two slopes you must get Negative 1. so if the product of the two slopes are negative one then they are perpendicular so this is the thing now we will see problem over here determine whether line and line CD are parallel perpendicular neither four the point is three comma 6 is at minus 9 comma two is at 5 comma 4 and is at 2 comma three graph each line to verify your answer so first we need to graph that is point is that how do we graph this whereas this this is the axis the vertic horizontal and the vertical over here the scale is 2 4 6 so 3 is somewhere over here and the 6 is over here so this is the first point is minus 9 over here and 2 is over here so and then draw straight line between those two points same way for is Phi that is somewhere over here and 4 so this is the point and is 2 and 3 that is over here draw straight line you will get the two lines so looking at this you can tell it cannot be perpendicular it should be parallel or not parallel so you can you know eliminate this completely you can eliminate this completely you can just tell whether this or this now let us see how do we exactly find it use the soap slope formula and find it you know the two points so this will be X1 y1 2 2. so now use the formula Y2 minus y1 divided by 2 minus 1 and find it so Y2 is 2 minus 6. same with minus 9 minus minus 3. put it in the calculator you'll get the answer 1 by 3. now same thing do it for the other line is other points 1 1 2 2 and you will get the answer so over here it is 2 is 3 minus 4 divided by 2 minus 5 put in the calculator you will be getting the Answer 1 by 3 and 1 by 3. here you can see the slopes are same therefore they are parallel lines so this is how we get it so I'm sorry this is not Pines it's lines just realized it okay it's parallel lines anyways now let's go to the next problem determine line relationship when given graphs the graphs are given over here in this case so in this previous one the graphs were not given we had been given points now let us see they are asked us to find whether RS and and are perpendicular parallel or neither so obviously you can rule out parallel because they're intersecting this is not true it should be the perpendicular or neither let's see them and the points are there see the points and same thing just like what we have done before you need to use the slope formula let's first write the points for line RS the Points are minus eight minus 3 is six and seven for minus 6 my minus 4 6 0 0. now use the slope formula and write it X1 y1 2 2 just substitute and solve over here you will get Y2 minus y1 Y2 is 7 minus of minus 3 so use the bracket divided by 6 minus of minus 8 equals to over here it'll be 10 and velb this minus minus becomes plus so it'll be add them all 10 by 14 it's 5 by 7. same way find the slope over here for the next line 0 minus 0 minus 6 and 0 minus of minus 4 will be negative 6 by 4 it's minus 3 by 2. it's not over now you need to multiply these both slopes you got the slopes multiply them now when you multiply over here you're getting the answer minus 1.07 this is not perpendicular if it was to be perpendicular you had to get exactly -1 but this is not minus one it's minus 1.07 so they are not perpendicular so that neither basically they're neither parallel nor perpendicular so same way let's do one more problem over here again the graphs are given the line EF and DG so same like what we have done before so what would ask you now is please pause the video try this by yourself and then continue with the explanation together hope at least some of you have tried to you know try this problem but now we will just quickly look at this write the given points and write the given points over here then use the slope formula Y2 minus y1 so we get this is X1 y1 2 2 just apply over here minus 1 minus 6 divided by 6 minus 3 and write the answer same way slope of DG will be 5 minus of minus 1 to all minus of minus 2. get the answers now over here all you need to do is multiply the slopes multiply them when you multiply them we are getting exactly minus one how because the 7 7 cancels over numerated denominator three three cancels only negative 1 remains therefore they are perpendicular lines so this is the the so these are perpendicular lines that's the answer
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