Distance between points in a plane

hello and welcome in this video we will learn how to find the distance between two points using the formula the distance between a pair of points PQ it’s given by the difference in the x coordinates squared plus the difference of the y-coordinate squared on the route difference of Y coordinates squared under all this under root no the distance of a point P x1 y1 from the origin origin has coordinate 0 and 0 is the difference of the x-coordinate plus squared minus plus the difference of the y-coordinates squared under root which gives you the x coordinate squared plus y coordinate square of the point p on the root now the distance between two points lying on x-axis if a point is lying on x axis its y coordinate is zero difference of x coordinate squared plus the difference of y coordinate squared under root which gives you difference of the x coordinate squared on the road which is equal to the absolute value of the difference of the x coordinates similarly if the points are lying on the y axis they’re x coordinate is zero in this case the distance PQ would be a difference of the x-coordinate squared which is 0-0 plus the difference of the

y-coordinates squared on the root gives you in the absolute value of the difference of the Y coordinates let us find the distance between a pair of points negative 3 1 and 3 comma 2 using the formula that we just discussed let these points B P negative 3 1 and q 3 and to let these be x1 y1 x2 y2 now using the formula the difference of the x-coordinates negative 3 and 3 squared plus the difference of Y coordinates squared on the route so that gives you negative 6 squared is positive 36 plus negative 1 squared is positive 1 under root gives you Drew’s 37 so that’s the distance between these two points let us take another example and find the distance between points P a cos alpha a sine alpha and cue a cos beta a sine beta just call this x1 y1 x2 y2 now let us apply the formula P Q equals to the difference of the x coordinates which is a cos alpha minus a cos beta squared plus the difference of the Y coordinates a sine alpha minus a sine beta squared under root let us expand this squared using the algebraic formula a plus B squared equals 2 a squared plus B squared plus 2a be here it is a minus B squared which is a square minus 2 a B plus B squared so let us apply the same formula here thank you sue a squared cos squared alpha plus a squared or squared beta minus 2 a squared cos alpha cos beta that’s the expansion of the first parentheses squared plus a squared sine squared alpha plus a squared sine squared beta minus 2a squared sine alpha sine beta that’s expansion of the second parentheses squared all this under root all this alert now let us try to pair this and use to go metric identities to simplify let us simplify some of these element comparing a squared cos squared alpha

with a squared sine squared alpha because I can use sine square alpha plus cos squared alpha equals to one identity of the trigonometry and plus I’m pairing and beat us now a square cos square beta a square or squared beta plus a squared sine squared beta and I’m paying these together ok coz alpha cos beta minus 2 a squared sine alpha sine beta all this under fruit now let us factor out the a squared when squared alpha plus cos squared alpha or this plus a squared so in squared beta plus R squared beta for that part and also factored out to a squared negative 2a squared which gives you cos alpha cos beta plus sine alpha so in beta all this under root now you have here a squared this is one Plus this again he is one that is a squared minus 2 a squared because alpha cos beta plus sine alpha sine beta that’s the remaining term now let us see what formulas we have used here we have used the technologic identity sine squared theta plus cos squared theta equals to 1 we use that so we have a squared plus a squared minus 2a squared cos alpha cos beta plus sine alpha sine beta okay on this in the root and here we’ll be using another identity for cosine Phi cos beta plus sine alpha sine beta which is let us use the identity directly here a squared plus a squared is 2a squared minus 2x squared and for this will use the identity cos alpha minus beta that is the identity we used to replace cos alpha cos beta plus sine alpha sine beta and now we can factor out the 2a

squared 1 minus cos alpha minus beta under root I guess here 2a squared and this can be written as two sine squared alpha minus beta over to under it thank you 4a squared sine squared alpha minus beta over 2 under root you may want to refer to the trigonometric identities video for this part of the problem because it requires knowledge of total metric identities so 4a squared would be 2a and sine squared alpha minus beta on the road will be sine alpha minus beta over 2 so that’s the distance between these two points now let us take another example find the value of the a if the distance between the points a comma 2 and 3 comma 4 is 2 root 2 now let us use the formula here this is the x coordinate y 1 second x coordinate and the second y coordinate now the distance let these points be P and Q is the difference of the x coordinates which is a minus 3 squared plus the difference of the y-coordinate square under root but the distance is already given as 2 root 2 so let us equate this to 2 root 2 PQ now is to vote two equals 2 as expand it first Francis a squared minus six a plus 9 plus 2 minus 4 is negative 2 squared is 4 on the road a squid minus XA plus 13 one dessert 22 now if we square on both sides you will have four times two equals 2a squared minus 6ei + 13 eight equal to a squared minus 6 a plus 13 now let us solve this a squared minus 6 a plus 13 equals 28 which gives you a squared on a sexy plus 5 equals to zero so I’m adding negative 8 on both sides or I’m subtracting 8 on from both sides now we’ll now we have a quadratic equation we can use the factoring method and solve this a square the factors of five are five and one so I’ll take

now is weird I will take a out of this term gives me a minus 5 minus a out when this one out actually a minus 5 plus 40 I’m just factoring out a from the first term and negative one from the second now i have e minus five factored out and you are left with a minus one so we have a is equal to five or a equals to 1 as the values of a thank you for watching