[Music] [Music] okay so in this video we want to have a look at perpendicular and parallel lines let's focus on parallel first so that's a little bit easier so if we have parallel lines um and we use arrows on two straight lines to indicate that they are parallel um and if they are if a b is parallel to c d this is how we write it a b is and then we have those two vertical lines that indicate that it's parallel to cd what can we say about the gradients of parallel lines okay so we know that parallel lines have same gradient so let's have a look at the example where we might need to refer to the lines being parallel show that the line passing through the points 6 4 and 7 11 is parallel to the line passing through the points 0 0 and 1 7 okay so we want to show that so we need to be able to show that we understand what gradient means and we understand what it means for two lines to be parallel okay it's not up for the person to the person marking our work to be able to intuit things from our working it's up to us to show it explicitly be really clear about the fact that we understand what it means for two lines to be parallel so the first thing we're going to need to do is calculate the respective gradients and obviously if they're parallel we're going to expect them to be the same but when they are the same we then need to draw a conclusion about that therefore because the gradient of a b is the same as the gradient of p q the lines a b and p q must be parallel okay so let's work out the gradient of a b so referring back a few videos ago to calculating gradient rise over run changing the y values so 11 minus four over changing the x values being consistent about how you use the coordinates so i need to be seven minus six eleven minus four is seven seven minus six is one and so the gradient from a to b is seven separately we'll work out the gradient from p to q so you don't just say the gradient from p to q is 7 so therefore the lines are parallel you actually need to show that pq also has a gradient of 7 and then draw a conclusion so again changing the y values over changing the x values so 7 over 1 which is also 7. so therefore since the gradient of a b is equal to the gradient of p q a b must be parallel to p q okay all right perpendicular lines if a b is a perpendicular means that the lines are at right angles so there's a 90 degree angle in between them okay um if line a b is perpendicular to c d we write this a b is it's sort of like a little upside down t perpendicular to c d okay there's lots of um different sort of geometries that we can do here but one of the things that's important is that if this is a right angle in between them let's just think about two triangles so let's think about this i'm just going to draw it slightly color-coded reasons that may or may not appear obvious okay so let's think about this purple triangle here versus this blue triangle here okay so we know that they're both right angle triangles obviously i can you can see because of the scale that they're the same triangle the diagram's a little bit skewed but you can see from the grid there you know the red length is one and the green length is two so therefore they're the same triangle um but equally we can also kind of work that out from the 90 degree angle so um we know that if we call if we were to call let's say if we were to call that angle um theta for example then that angle up there would be 90 minus theta okay because of the sum of the angles in the triangle 180 degrees you've got 90 degree right angle already here so the other two angles have to add up to 90 degrees now we also know that that angle in there is 90 minus theta okay that's the same thing here and then we know that this angle over here this one up here is going to be the same as this one down here because they're vertically opposite angles so that's also 90 minus theta which means that this angle up here must be theta okay so what you have now is two these two triangles have exactly the same angles and so therefore they must be similar triangles but we can also show that they've got exactly the same lengths as well so if they're exactly the same triangle now i know you could probably say i could tell they were the same by looking at them but let's sort of we're trying to be a bit more precise than that okay so if they're exactly the same triangles the lengths are all the same okay so we know and we know they're the same um so we know that that length there let's call that i know that that's one on this scale but let's just recall that generally the run and let's call this generally the rise so if you think about this other triangle which we know is the same a congruent triangle and this becomes the run so we know that the gradient of a b is rise over run okay we know that maybe i won't actually maybe i won't call it rise and run because i'm going to confuse you let me just change the notation there let's just call it two different lengths let's just call that um you know x and that length y that's a bit confusing as well but i'll go with capital x and capital y let's call that capital x and that capital y okay so if we're working out the gradient of a b we know that that is rise over run in this case that's going to be y over x okay whereas when we look at the gradient of c d we can see that that clearly has a negative gradient and that its rise is actually x and its run is actually y okay so this connection from this to this is important when we look at perpendicular lines okay so we can see that what is the rise in the run switches around so in terms of calculating gradient we need to flip our fraction we call that taking the reciprocal okay so when we reverse the numerator and denominator we've taken the reciprocal of the fraction and we also need to make it negative so the perpendicular gradient is the negative reciprocal now we can verify that for this line because we can work out that rise here is 2 and run here is 1 so the gradient of that is 2 over 1 which is 2 whereas the gradient of this is clearly negative and the rise is 1 and the run is 2. so gradient is 2 gradient is negative half there are two ways to look at the relationship between the gradients of perpendicular lines one is that if you multiply the gradients of two perpendicular lines together you should get negative one we can see that up here in this example negative half times positive two will give us negative one um or the other alternative and the more useful practical one is this idea of the negative reciprocal gradient two will be the negative reciprocals we've flipped gradient one upside down of gradient of um yeah so gradient two will be the negative reciprocal of gradient one okay so let's just have a look at that idea so state the gradient of a line that is perpendicular to a line with gradient negative two okay so if a line has a gradient of negative two if it helps you can think about that as negative two over one so the perpendicular gradient is going to be equal to positive because we multiply by negative one so that'll change it to a positive and we flip it so it's going to be equal to positive one-half okay if the original gradient was one-quarter the perpendicular gradient is going to be we multiply by negative one so it was positive up here so it's now going to be negative okay and we flip it if we flip one over four we get four over one and four over one is just four okay if the gradient was one one is one over one the perpendicular gradient is going to be equal to now we met we multiply it by negative one so it's going to be it was originally positive it's now going to be negative and if we flip it we still get one over one and so the perpendicular gradient is negative one if the original gradient was negative two-thirds the perpendicular gradient we're going to multiply by negative one so it's going to become positive okay and we flip the two thirds so it becomes positive sorry positive three on two okay consider the points a b p and q we've got coordinates there q has an unknown x coordinate find t if a b is perpendicular to p q okay let's calculate the gradient of a b gradient of a b is 12 minus zero over zero minus six so that's 12 over negative 6 which is negative 2 okay gradient of pq is um i'm going to do it i'm going to so i have the a bit easier equation to solve i want to have i want the denominator to be t minus eight rather than eight minus t just makes the equation a bit nicer but you'll get there in the long run if you do it the other way around so if i want t minus 8 on the denominator i need to have 8 minus 10 on the numerator so it's negative 2 on t minus 8. now perpendicular gradient to this perpendicular to negative 2 is positive one-half and so therefore we need the gradient of pq to be equal to positive one-half and so we can solve that equation okay so let's multiply both sides of the equation by two well we can do it in two ways multicross multiply which i don't like because i find students don't understand what they're doing and they often use it incorrectly but in this situation when you've got one fraction equal to another fraction you can cross multiply so you get negative four equals one times t minus 8. okay that's the quickest way to do it if you need to the failsafe method common denominator your common denominator in this instance is going to be 2 lots of t minus 8. so if we did that 2 lots of t minus 8 you've multiplied the denominator of the left hand side by 2 so you do the same to the numerator you've multiplied the denominator of the right hand side by t minus 8 so you do the same to the numerator now you can multiply by your denominators and you get straight to the equation that we had here okay so when you're cross multiplying that is what's happening okay but you can only cross multiply if you have one fraction equal to another fraction if there's nothing else in the equation the minute there's something else um you need to um you're going to need to use another method um okay so negative 4 equals t minus 8 adding 8 to both sides we find that t equals positive 4. so if t equals positive 4 if we stick that back up there we're going to have negative 2 and negative 4 so we're going to have a gradient of positive a half which is indeed perpendicular to negative 2. okay question four find the equation of the line which is parallel to this line and which passes through this point okay so parallel to this line is telling us about gradient so we need to work out what the gradient of this line is and then because they're parallel our line will have the same gradient so we first of all need to get this line into gradient intercept form so we know we have four x plus two y equals one okay we want to make y the subject so 2y equals negative 4x plus 1 dividing by 2 y equals negative 2x plus half okay so the gradient is negative 2. our line is parallel and so the gradient we want is also going to be negative 2. okay now so we've got a gradient of negative 2 going through the point 3 5 and so now it's about finding the equation of a line as you know i prefer to use y minus y1 equals m times x minus x1 i substitute in my point the y coordinate of my point gradient and the x coordinate of my point and then i rearrange to make y the subject adding 5 so minus 2x plus 11. or you can use y equals mx plus c you sub in the gradient which is negative 2 then you sub in your point which means when x equals three y equals five c equals eleven and then you put c back into your original equation and you get your final answer okay personally i find this is quicker and there's less steps it's a bit more streamlined you put all the information in you make y the subject this will always work i find students are always drawn to y equals mx plus c even though it's not the most efficient have a play with the other version see if when once it becomes familiar with it you become more more comfortable okay question five find the equation of the line which is perpendicular to this line okay so this is giving us our information about gradient and passes through this point okay so perpendicular that line we want to think about that what the gradient of that line is so it's y equals two x plus one all over five let's be clear that that means that's the same as two x on five plus one on five okay and so here's our gradient here two-fifths now our line is perpendicular to this line so our gradient is going to be negative 5 on 2 okay so that's what we need and we've got our point so i'm going to find the equation i'm not going to do it twice this time if you want to do y equals mx plus c by all means have a pause and do y equals mx plus c and then let's see if we end up we should if we've both done it correctly end up with the same answer at the end okay so it's y minus the y coordinate equals the gradient times x minus the x coordinate so expanding the brackets minus 5 on 2x plus 5 on 2 and then we want to add 1 1 is the same as 2 on 2 so 5 on 2 plus 2 on 2 is 7 on 2. okay question six find the equation of the perpendicular bisector of the line segment a b where a is seven zero and b is one eight now i should have left us some more room here i'm just going to draw a rough sketch off to the edge here because i want to explain what a perpendicular bisector is so we need one eight and seven zero okay so let's put seven zero there that's gonna be that's a and one eight somewhere up here which is b okay so we want to find the perpendicular bisector so perpendicular we've talked about it's going to be a line that's at right angles to this but it also needs to bisect which means cut in two okay and actually needs to cut it evenly into so it's going to need to be both perpendicular and going through the midpoint okay so we want to work out we need to find gradient and a point to find the equation of the red line so we're going to need to find the gradient of a b and use that gradient to find the gradient of the red line but we're also going to need to use the midpoint as the point we can't use either of these points the perpendicular bisector doesn't go through either of these points okay so let's calculate our gradient of a b first gradient of a b rise over run you've got your diagram if you need eight minus zero over one minus seven so that's eight over negative six so negative four thirds okay so that means that the perpendicular gradient is going to be positive three-quarters okay and then we also need the midpoint yes sorry i've left us a bit crowded here the midpoint remember is the average of the x-coordinates seven plus one over 2 and the average of the y coordinate 0 plus 8 over 2. so that's 8 over 2 and over 2 so 4 4. okay so we've got our gradient and our point and so we can find our equation of the perpendicular bisector so it's going to be y minus the y-coordinate equals the gradient times x minus the x-coordinate expanding out the brackets on the right three-quarters x three-quarters times four please try try to get past writing that thinking about that as 12 over four three divided by four and then times four the fours are cancelling out that's going to cancel with that when you multiply them together okay so we're just gonna get minus 3 and then we need to add 4 to both sides so 3 quarters x plus 1 is the equation of our perpendicular bisector and that vaguely makes sense i haven't drawn scale so you know this should be a bit higher up but things aren't perfectly scaled but it's making sense a positive gradient definitely and a y intercept somewhere near the origin is making sense okay once again there are some much older videos explaining these concepts linked there and the practice work for today is from exercise one i think that's one eye not one l and one eye on page 59 of your textbook and those are the questions you