in this video we're going to go over some problems dealing with coordinate geometry so in this example what are the coordinates of point b so notice that we have a vertical line which means that point b and point a share the same x coordinate which is one now here we have a horizontal line which means that point b and c share the same y coordinate which is two and so that's how you can find the x and y coordinate of point b so whenever you have a vertical line the x coordinates will be the same if you have a horizontal line the y coordinates will be the same so now what is the area of the triangle to calculate the area of a right triangle it's one half base times height so the base is the distance between points b and c and the height is the distance between points a and b so what's the distance between points b and c notice that the y values are the same so the distance between the two points is simply the difference between the x values so five minus one is four so the base of the right triangle is four units now the height notice that the x values are the same so we'll have to worry about that the difference is five minus two and so the distance between a and b is going to be three and now we can calculate the area so it's going to be one half base times height the base is for the highest stream half of four is two and two times three is six so the area of this right triangle is six square units the endpoints of a diameter of a circle are one comma two and seven comma ten what are the coordinates of the center of the circle so let's say this is the first point one comma two and over here this is uh seven ten so let's say the center is somewhere in between so how can we identify the coordinates of the center of the circle let's call this a and b now what we need to do is we need to use the midpoint formula and so the midpoint of a and b will give us the center c and the midpoint formula is the average of the x values and the y coordinate will be the average of the y values so let's call this x one and y one and this is going to be x two and y two so we need to average one and seven the midpoint of one and seven is four and the midpoint of two and ten the middle number is six so the center should be four sixths but if you want to use the formula it's going to be one plus seven divided by two and for the y values two plus ten divided by two so one plus seven is eight and two plus ten is twelve now eight divided by two is four and twelve divided by two is six so that's how you can use the midpoint formula to get the midpoint between two points and so the center of the circle is four comma six so that's the answer for part a now what about part b what is the area and circumference of the circle in order to calculate that we need to calculate the radius of the circle first so how can we calculate the radius of the circle well we need to use the distance formula and here it is it's x2 minus x1 squared plus y2 minus y1 squared so let's say this is x2 and y2 and let's see the center is going to be x1 and y1 so in this example x2 is 7 x1 is 4. y2 is 10 y1 is six now seven minus four is three and ten minus six is four three squared is nine four squared is sixteen nine plus sixteen is twenty-five and the square root of 25 is 5. so the distance between points c and b is 5 which means that the radius of the circle is 5. so now that we have the radius of the circle we can calculate the circumference and the area of the circle so the area of the circle is simply pi r squared so that's going to be pi times 5 squared so that's 25 pi the circumference is 2 pi r so that's going to be 2 pi times 5 and so that's 10 pi and so that's how you can calculate the area and the circumference of this particular circle now what about part c what is the standard equation that describes the graph of the circle for a circle the standard equation is x minus h squared plus y minus k squared and that's equal to r squared and so the center is h comma k so notice that h is four and k is six and we have the radius r so it's going to be x minus four squared plus y minus six squared and that's equal to five squared which is twenty-five so this is the standard equation of the circle and so that's the answer number three given the coordinates of a and b shown below what is the equation of the tangent line that touches circle a at point b now since we're dealing with circle a a is the center of the circle and b is on the edge of the circle which means that segment a b represents the radius of the circle so anytime you draw the radius to a tangent line it always meets the tangent line at right angles so if we could find the line between a and b or the equation of the line between a and b we can write the equation of the tangent line which let's call the tangent line l so how can we write the equation of the line that passes through points a and b first we need to calculate the slope so this is point a three comma two and point b which is five comma eight let's call this x one y one and x two and y two the slope is the rise over the run it's equal to y2 minus y1 divided by x2 minus x1 is the change in y divided by the change in x so notice that the run going from 3 to 5 is 2. the rise going from 2 to 8 is 6. so the slope is rise over run represented by the symbol m so it's 6 over 2 which is 3. if you use the formula y 2 is 8 and y 1 is 2 x 2 is 5 x 1 is 3 8 minus 2 is 6 5 minus 3 is 2 and you get the same answer so the slope is equal to 3. so now that we have the slope we can write the equation of the line now we can use the slope intercept formula or we can use the point slope formula there's many ways to do this but i'm going to use the point slope formula so y one is two m is three x one is three based on those values you can use these values too you'll get the same equation so i'm going to distribute the 3 so it's going to be 3x minus 9 and then i'm going to add 2 to both sides negative nine plus two is negative seven so this represents the equation of the line that passes through points a and b now let's write the equation of the tangent line first we need to calculate the slope of the perpendicular line or the tangent line the tangent line is perpendicular to line a b and so what is the slope of a perpendicular line to find it what you need to do is flip the fraction and change the sign so instead of being three over one it's going to be one over three and instead of being positive three it's going to be a negative one over three so that's how you can find a slope of a perpendicular line keep in mind the slope of two parallel lines is the same but for a perpendicular line is the negative reciprocal of the first line so this is the slope of the tangent line so how can we write an equation given the slope and the point we can't use this point because a is not on a tangent line we have to use point b that's on the tangent line so if you're given a slope and a point you could use the point slope formula like we did before or you could use the slope intercept formula both will work this time i'm going to use the slope and a sub formula so you could see how to do it using both methods so i'm going to replace y with eight and m is negative one third and i'm going to replace x with five now what i need to do is solve for b so i'm going to get rid of this fraction by multiplying everything by three so eight times three is 24 negative one third times three that's negative one times five so that's just negative five and then three times b is three b so now let's add five to both sides so 29 is equal to three b so now let's divide both sides by three so b is 29 over three so now we can write the linear equation so first let's start with the slope intercept form all we need to do is replace m and b so it's going to be negative 1 3 x plus 29 over 3. so this is the equation of the tangent line in slope intercept form now if you want to put it in standard form you can do this let's say if you want to get it in this form a x plus b y is equal to c let's multiply everything by three so it's going to be three y that's equal to negative one x plus 29 and then i can move this to the side so i have 1x plus 3y is equal to 29. so you can write the equation of the tangent line like this if you want to there's many ways in which you can write it and so that's it for this problem number four what is the area of the region bounded by the x-axis the y-axis and the graph for x plus y is equal to 8. so let's draw a picture now what we need to do is calculate the x and the y-intercepts first so let's start with the y-intercept how can we calculate the y-intercept of this equation in order to find the coordinates of the y-intercept replace x with 0 and solve for y so 4 times 0 is 0 and so we can see that y is equal to eight so the y intercept is zero comma eight because we replace x with four now let's calculate the x intercept so this time we're going to replace y with zero so we have four x is equal to eight and now let's divide both sides by four so x is two so the x intercept is two comma zero so now let's graph it so let's say this is 2 and this is 8. so therefore we have a graph that looks like this let's do that one more time there we go so we need to calculate the area of the region bounded by the y-axis the x-axis and this graph so basically we need to calculate the area of this triangle so we have a right triangle with a base of two and a height of eight so the area of this triangle is one half base times height so it's half times two times eight half of two is one one times eight is eight so that's the area of the shader region it's eight square units number five plot the point 3 comma 4 comma 5. so let's start with that first so let's say this is the x-axis this is the y-axis and this is the z-axis so first let's travel three units to the right and then let's travel four units along the y-axis so let me draw a line that's parallel to the y-axis and also one that's parallel to the x-axis so right now that's at x equals three and y equals four and then we need to go up five units i'm just going to draw a solid line so our point of interest is right here so we traveled three units in the x direction and then four units in the y direction and then five units in the positive z direction so there it is so that's three comma four comma five i know my graph is not perfect but you can make the best of it so how can we calculate the distance between the origin and point p so what is the distance between those two points now let's call the origin point a which is zero zero zero there's a simple way in which you can calculate the distance between two points in three dimensions so we have point a and point p so this is x1 y1 and z1 this is x2 y2 and z2 so the distance between two points is simply going to be x2 minus x1 squared y2 minus y1 squared plus z2 minus z1 squared all within a square root symbol so we can see that x2 in this example is 3 x1 is 0. y2 is 4 y1 is 0. z2 is 5 z1 is zero so what we have is 3 squared plus four squared plus five squared so three squared is nine four squared is sixteen five squared is twenty-five nine plus sixteen is twenty-five and twenty-five plus 25 is 50. so we need to simplify the square root of 50 which is 25 times 2 and the square root of 25 is 5. so the distance between the origin and point p is five square root two and so as a decimal that's about seven point zero seven approximately number six what is the distance between the point five comma three and the line three x plus four y minus seven equals zero so let's say if this is the line and here is the point the distance is the perpendicular distance between the line and the point so we need to calculate this distance d and it's a simple formula in which we could do so so this is going to be x1 and this is y1 and it's in the form ax plus by plus c is equal to zero so here's the formula that you need so the distance is going to be the absolute value of ax one plus b y one plus c divided by the square root of a squared plus b squared so in this example a is three x one is five b is four y one is three and c is negative seven and then a squared that's three squared b squared is four squared now three times five is fifteen four times three is twelve 3 squared is 9 4 squared is 16. now 15 plus 12 that's 27 and 9 plus 16 is 25. now 27 minus 7 is 20 and the square root of 25 is 5. now 20 divided by 5 is 4. so the distance between the point and the line is 4 units long seven a b c d is a square what is the area of the shaded region if the x intercept of the circle is four comma zero so this is four comma zero which means this part here is negative four comma zero and the y intercept is zero comma four and the other y intercept is zero negative four now to calculate the area of the shaded region it's going to be the area of the square minus the area of the circle because the square is outside of the circle so it's the area of the large figure minus the area of the smaller figure now what is the area of the square so the radius of the circle we can clearly see that it's 4 units so the diameter is 2r 2 times 4 is 8. and so this also represents the radius of the circle so this side is also eight now if you have a square where every side is eight the area of the square is simply s squared and the area of the circle is pi r squared so the side length of the square is eight and the radius of the circle we can see that it's four so it's going to be eight squared which is 64 minus pi times four squared or 16 pi so this is the exact answer for the area of the shaded region now if you want to get a decimal value you can convert 16 pi into a decimal which is about 50.2655 and so the area as a decimal is about 13.7345 square units so you can report your answer like that if you want to but this represents the exact answer number eight triangle abc is an equilateral triangle and point c is a comma zero what is the area of the triangle so if this is eight comma zero we could say that ac is eight and for an equilateral triangle all three sides are the same so to calculate the area of an equilateral triangle is the square root of three divided by four times the side lamp squared so s in this example is eight and eight squared is 64. and 64 divided by 4 is 16. so the area of the equilateral triangle is 16 square root 3 square units and so that's it for part a now let's move on to part b what are the coordinates of point b so point a is clearly zero zero what about point b now for an equilateral triangle because all three sides are the same the three angles are congruent so 180 divided by 3 is 60. so each angle has a measure of 60 degrees now what i'm going to do is draw a line and split the triangle into two triangles so this is going to be 90 degrees this is going to be 60 and half of 60 is 30. so this angle is 30. and so the hypotenuse is 8. let's call this new point point d now if we have a 30-60-90 triangle what are some things that we need to know across the hypotenuse is two across the side from the 30 degree angle is going to be half of whatever the hypotenuse is so this is 1. across the 6 degree angle is going to be whatever this is times the square root of 3. so the hypotenuse is 8. so across the 30 degree angle it's going to be half of 8 which is 4. so this side is 4 that side is 4. across the 60 degree angle it's going to be whatever this is times the square root of 3. so bd is 4 square root 3. so now we have the x coordinate of point b it's 4 and the y coordinate is 4 square root 3. so the coordinates are 4 comma 4 square root 3. so that's the answer number nine the three vertices of a triangle are one comma two five comma ten and seven comma four write the equation of the median to segment ac so let's draw a picture so let's say this is the triangle and this is point a b and c so first we need to write the equation of the median to segment ac so the median of ac is going to be the midpoint of ac and so the median is the segment that extends from the opposite vertex from vertex b to the midpoint of ac so it splits ac into two congruent parts so bm is the median and m is the midpoint of ac so how can we write the equation of this line to write the equation of any line you either need the two points that are on the line or a point in a slope we have point b it's five comma ten and we don't have the midpoint but we can find the midpoint of ac using the midpoint formula so the midpoint formula is x1 plus x2 divided by 2 and then y1 plus y2 divided by 2. so let's call this x1 and y1 and this is going to be x2 and y2 so it's 1 plus 7 divided by 2 and 2 plus 4 divided by 2. so one plus seven is eight two plus four is six eight divided by two is four six divided by two is three so the midpoint is four comma three now let's calculate the slope of segment bm so i'm going to call this x2 and y2 and this one is going to be x1 and y1 so the slope is the change in y divided by the change in x so y two is ten y one is three x two is five x one is four ten minus three is seven five minus four is one so the slope of that line is seven now let's use the point slope formula y minus y one is equal to m times x minus x one so let's use this point y one is three the slope is seven based on what we have here and x one is four so let's distribute the seven so it's gonna be seven 7x minus 28 and then let's add 3 to both sides so y is equal to 7x minus 25 and so this is the equation of the median so now let's work on our next example and that is part b so we're going to use the same triangle triangle abc and this time what we need to do is we need to write the equation of the perpendicular bisector of ac so there's two things that we need to keep in mind with the perpendicular bisector it bisects the segment into two congruent parts so that means that it touches the midpoint of segment ac and we said the coordinates of the midpoint were four comma three so we're going to need that and then let's draw a perpendicular line that passes through this point now this is the perpendicular bisector it meets ac at a right angle so it's 90 degrees and it bisects ac into two congruent parts the perpendicular bisector does not have to pass through vertex b so just keep that in mind so we cannot use point b when writing the equation of the perpendicular bisector the only way it's going to pass through point b is if we have like an equilateral triangle or something so we need to use point m now what is the slope of the perpendicular bisector let's call this line l what is the slope of line l now first we need to calculate the slope of line ac then we could find it for line l so a is one comma two and c is seven four so let's call this x1 y1 x2 and y2 so the slope for segment ac is going to be y2 minus y1 divided by x2 minus x1 so that's going to be 4 minus 2 over seven minus one four minus two is two seven minus one is six and two over six reduces to one over three so that's the slope of segment ac now the slope of the perpendicular line is going to be the negative reciprocal of that so it's going to be 3 over 1 but negative and so that's the slope of the perpendicular bisector now let's write the equation of the line using the point and the slope so this time i'm going to use the point slope formula for those of you who prefer that method so y is going to be 3 and x is 4 and m is negative 3. so let's calculate b negative 3 times 4 is negative 12. and then let's add 12 to both sides so 3 plus 12 is 15. so the y intercept is 15. so now that we have the value of b we can write the equation all we need to do is replace m and b so it's going to be y is equal to m which is negative 3 and b is 15. so this right here is the equation of the perpendicular bisector of ac now let's move on to the last part part c so what we need to do is write the equation of the altitude so once again let's say this is a b and c so the altitude of ac is going to pass from vertex b to ac and it's going to be at a right angle so this is not going to be the midpoint anymore so this time we need to use point b which is 5 comma 10. and it's perpendicular so the slope should be the same the slope of the perpendicular bisector we said was negative three and so that's not going to change the only difference is we have a new point so let's use the point slope formula y minus y1 is equal to m times x minus x1 this is perpendicular to ac by the way so y1 is going to be 10 x1 is going to be 5. so let's distribute the negative 3. so it's going to be negative 3x plus 15. and then let's add 10 to both sides so y is going to equal negative 3 x plus 25 and so this is the equation of the altitude of ac so the perpendicular bisector and the altitude they share the same slope because they're both perpendicular to ac so we don't have to recalculate this value and so this is the answer you