Key Problems in Coordinate Geometry

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 1. Now here we have a horizontal line, which means that point B and C share the same y-coordinate. which is 2. 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 1 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 C. 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 5 minus 1 is 4. So the base of the right triangle is 4 units. Now, the height, notice that the x values are the same, so we don't have to worry about that. The difference is 5 minus 2. And so the distance between A and B is going to be 3. And now we can calculate the area. So it's going to be 1 half base times height. The base is 4, the height is 3. Half of 4 is 2. And 2 times 3 is 6. So the area of this right triangle is 6 square units. The endpoints of a diameter of a circle are 1, 2 and 7, 10. What are the coordinates of the center of the circle? So let's say this is the first point, 1, 2. And over here this is 7, 10. So let's say the center is somewhere in between. So how can we identify the coordinates of the center of a 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 x1 and y1 and this is going to be x2 and y2. So we'll need to average 1. and 7. The midpoint of 1 and 7 is 4. And the midpoint of 2 and 10, the middle number is 6. So the center should be 4, 6. But if you want to use the formula, it's going to be 1 plus 7 divided by 2. And for the y values, 2 plus 10 divided by 2. So 1 plus 7 is 8. And 2 plus 10 is 12. Now 8 divided by 2 is 4. And 12 divided by 2 is 6. So that's how you can can use the midpoint formula to get the midpoint between two points. And so the center of the circle is 4 comma 6. 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 say the center is going to be x1 and y1. So in this example, x2 is 7, x1 is 4, y2 is 10, y1 is 6. Now 7 minus 4 is 3, and 10 minus 6 is 4. 3 squared is 9, 4 squared is 16. 9 plus 16 is 25, 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 4 and K is 6 and we have the radius R so it's going to be X minus 4 squared plus Y minus 6 squared and that's equal to 5 squared which is 25 so this is the standard equation of the circle And so that's the answer. Number 3. 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 AB 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, 3, 2, and point B, which is 5, 8. Let's call this X1, Y1, and X2, and Y2. The slope is the rise over the run. It's equal to Y1. minus y1 divided by X 2 minus X 1 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, y2 is 8 and y1 is 2, x2 is 5, x1 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 y1 is 2, m is 3. x1 is 3 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 9 plus 2 is negative 7 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 and 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 3 over 1, it's going to be 1 over 3. And instead of being positive 3, it's going to be negative 1 over 3. 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, it's 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 a 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-intercept formula, so you can see how to do it using both methods. So I'm going to replace y with 8, and m is negative 1 third. And I'm going to replace x with 5. Now what I need to do is solve for b. So I'm going to get rid of this fraction by multiplying everything by 3. So 8 times 3 is 24. Negative 1 third times 3. That's negative 1 times 5, so that's just negative 5. And then 3 times b is 3b. So now let's add 5 to both sides. So 29 is equal to 3b. So now let's divide both sides by 3. So b is 29 over 3. 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 third 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. ax plus by is equal to c. Let's multiply everything by 3. So it's going to be 3y, that's equal to negative 1x plus 29. And then I can move this to this 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 4x 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 8. So the y-intercept is 0, 8 because we replaced x with 4. Now let's calculate the x-intercept. So this time, we're going to replace y with 0. So we have 4x is equal to 8. And now let's divide both sides by 4. So x is 2. So the x-intercept is 2 common 0. 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 2 and a height of 8. So the area of this triangle is 1 half base times height. So it's half times 2 times 8. Half of 2 is 1. 1 times 8 is 8. So that's the area of the shaded region. It's 8 square units. Number 5. Plot the point 3, 4, 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 3 units to the right. And then let's travel 4 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 3 and y equals 4. And then we need to go up 5 units. I'm just going to draw a solid line. So our point of interest is right here. So we traveled 3 units in the x direction, and then 4 units in the y direction, and then 5 units in the positive z direction. So there it is. So that's 3, 4, 5. 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 0, 0, 0. 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 to be x2 minus x1 squared plus 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 0 So what we have is 3 squared plus 4 squared plus 5 squared. So 3 squared is 9, 4 squared is 16, 5 squared is 25. 9 plus 16 is 25. and 25 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 5 square root 2. And so, as a decimal, that's about... 7.07 approximately. Number 6. What is the distance between the point 5,3 and the line 3x plus 4y minus 7 equals 0? 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 there'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 0. So here's the formula that you need. So the distance is going to be the absolute value of ax1 plus by1 plus c divided by the square root of a squared plus b squared. So in this example, a is 3. x1 is 5. b. is 4, y1 is 3, and c is negative 7. And then a squared, that's 3 squared, b squared is 4 squared. Now 3 times 5 is 15, 4 times 3 is 12, 3 squared is 9, 4 squared is 16. Now 5 18 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. 7. ABCD is a square. What is the area of the shaded region if the x-intercept of the circle is 4,0? So this is 4,0, which means this part here is negative 4,0, and the y-intercept is 0,4. and the other y-ness up is 0, negative 4. 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 8. Now if you have a square, where every side is 8. 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 8, and the radius of the circle, we can see that it's 4. So it's going to be 8 squared, which is 64, minus pi times 4 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 8. Triangle ABC is an equilateral triangle, and point C is A, 0. What is the area of the triangle? So if this is 8, 0, we could say that AC is 8. And for an equilateral triangle, all three sides are the same. So to calculate the area of an equilateral triangle, it's the square root of 3 divided by the square root of 3. divided by 4 times the side length squared. So S in this example is 8. And 8 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 0, 0. But 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 2. Across the side from the 30 degree angle is going to be half of whatever the hypotenuse is. So this is 1. across the 60 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 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 9. The three vertices of a triangle are 1, 2, 5, 10, and 7, 4. 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 the slope. We have point B, it's 5, 10, 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 1 plus 7 is 8, 2 plus 4 is 6. 8 divided by 2 is 4, 6 divided by 2 is 3. So the midpoint is 4, 3. 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 y2 is 10, y1 is 3, x2 is 5, x1 is 4. 10 minus 3 is 7, 5 minus 4 is 1. So the slope of that line is 7. Now let's use the point slope formula. y minus y1 is equal to m times x minus x1. So let's use this point. y1 is 3, the slope is 7. we have here and x1 is 4. So let's distribute the 7 so it's going to be 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 4,3, 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 can find it for line L. So A is 1,2 and C... is 7, 4. 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 that's going to be 4 minus 2 over 7 minus 1. 4 minus 2 is 2, 7 minus 1 is 6, and 2 over 6 reduces to 1 over 3. 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 meet 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, 10. And it's perpendicular, so the slope should be the same. The slope of the perpendicular bisector we said was negative 3, 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 3x 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.

which is 2. 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 1 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 C. 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 5 minus 1 is 4. So the base of the right triangle is 4 units. Now, the height, notice that the x values are the same, so we don't have to worry about that. The difference is 5 minus 2. And so the distance between A and B is going to be 3. And now we can calculate the area.

So it's going to be 1 half base times height. The base is 4, the height is 3. Half of 4 is 2. And 2 times 3 is 6. So the area of this right triangle is 6 square units. The endpoints of a diameter of a circle are 1, 2 and 7, 10. What are the coordinates of the center of the circle?

So let's say this is the first point, 1, 2. And over here this is 7, 10. So let's say the center is somewhere in between. So how can we identify the coordinates of the center of a 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 x1 and y1 and this is going to be x2 and y2.

So we'll need to average 1. and 7. The midpoint of 1 and 7 is 4. And the midpoint of 2 and 10, the middle number is 6. So the center should be 4, 6. But if you want to use the formula, it's going to be 1 plus 7 divided by 2. And for the y values, 2 plus 10 divided by 2. So 1 plus 7 is 8. And 2 plus 10 is 12. Now 8 divided by 2 is 4. And 12 divided by 2 is 6. So that's how you can can use the midpoint formula to get the midpoint between two points. And so the center of the circle is 4 comma 6. 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 say the center is going to be x1 and y1. So in this example, x2 is 7, x1 is 4, y2 is 10, y1 is 6. Now 7 minus 4 is 3, and 10 minus 6 is 4. 3 squared is 9, 4 squared is 16. 9 plus 16 is 25, 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 4 and K is 6 and we have the radius R so it's going to be X minus 4 squared plus Y minus 6 squared and that's equal to 5 squared which is 25 so this is the standard equation of the circle And so that's the answer. Number 3. 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 AB 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, 3, 2, and point B, which is 5, 8. Let's call this X1, Y1, and X2, and Y2.

The slope is the rise over the run. It's equal to Y1. minus y1 divided by X 2 minus X 1 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, y2 is 8 and y1 is 2, x2 is 5, x1 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 y1 is 2, m is 3. x1 is 3 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 9 plus 2 is negative 7 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 and 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 3 over 1, it's going to be 1 over 3. And instead of being positive 3, it's going to be negative 1 over 3. 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, it's 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 a 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-intercept formula, so you can see how to do it using both methods.

So I'm going to replace y with 8, and m is negative 1 third. And I'm going to replace x with 5. Now what I need to do is solve for b. So I'm going to get rid of this fraction by multiplying everything by 3. So 8 times 3 is 24. Negative 1 third times 3. That's negative 1 times 5, so that's just negative 5. And then 3 times b is 3b.

So now let's add 5 to both sides. So 29 is equal to 3b. So now let's divide both sides by 3. So b is 29 over 3. 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 third 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. ax plus by is equal to c. Let's multiply everything by 3. So it's going to be 3y, that's equal to negative 1x plus 29. And then I can move this to this 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 4x 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 8. So the y-intercept is 0, 8 because we replaced x with 4. Now let's calculate the x-intercept.

So this time, we're going to replace y with 0. So we have 4x is equal to 8. And now let's divide both sides by 4. So x is 2. So the x-intercept is 2 common 0. 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 2 and a height of 8. So the area of this triangle is 1 half base times height.

So it's half times 2 times 8. Half of 2 is 1. 1 times 8 is 8. So that's the area of the shaded region. It's 8 square units. Number 5. Plot the point 3, 4, 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 3 units to the right. And then let's travel 4 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 3 and y equals 4. And then we need to go up 5 units. I'm just going to draw a solid line. So our point of interest is right here. So we traveled 3 units in the x direction, and then 4 units in the y direction, and then 5 units in the positive z direction. So there it is.

So that's 3, 4, 5. 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 0, 0, 0. 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 to be x2 minus x1 squared plus 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 0 So what we have is 3 squared plus 4 squared plus 5 squared. So 3 squared is 9, 4 squared is 16, 5 squared is 25. 9 plus 16 is 25. and 25 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 5 square root 2. And so, as a decimal, that's about... 7.07 approximately.

Number 6. What is the distance between the point 5,3 and the line 3x plus 4y minus 7 equals 0? 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 there'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 0. So here's the formula that you need.

So the distance is going to be the absolute value of ax1 plus by1 plus c divided by the square root of a squared plus b squared. So in this example, a is 3. x1 is 5. b. is 4, y1 is 3, and c is negative 7. And then a squared, that's 3 squared, b squared is 4 squared. Now 3 times 5 is 15, 4 times 3 is 12, 3 squared is 9, 4 squared is 16. Now 5 18 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.

7. ABCD is a square. What is the area of the shaded region if the x-intercept of the circle is 4,0? So this is 4,0, which means this part here is negative 4,0, and the y-intercept is 0,4.

and the other y-ness up is 0, negative 4. 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 8. Now if you have a square, where every side is 8. 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 8, and the radius of the circle, we can see that it's 4. So it's going to be 8 squared, which is 64, minus pi times 4 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 8. Triangle ABC is an equilateral triangle, and point C is A, 0. What is the area of the triangle?

So if this is 8, 0, we could say that AC is 8. And for an equilateral triangle, all three sides are the same. So to calculate the area of an equilateral triangle, it's the square root of 3 divided by the square root of 3. divided by 4 times the side length squared. So S in this example is 8. And 8 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 0, 0. But 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 2. Across the side from the 30 degree angle is going to be half of whatever the hypotenuse is. So this is 1. across the 60 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 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 9. The three vertices of a triangle are 1, 2, 5, 10, and 7, 4. 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 the slope. We have point B, it's 5, 10, 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 1 plus 7 is 8, 2 plus 4 is 6. 8 divided by 2 is 4, 6 divided by 2 is 3. So the midpoint is 4, 3. 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 y2 is 10, y1 is 3, x2 is 5, x1 is 4. 10 minus 3 is 7, 5 minus 4 is 1. So the slope of that line is 7. Now let's use the point slope formula. y minus y1 is equal to m times x minus x1.

So let's use this point. y1 is 3, the slope is 7. we have here and x1 is 4. So let's distribute the 7 so it's going to be 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 4,3, 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 can find it for line L. So A is 1,2 and C... is 7, 4. 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 that's going to be 4 minus 2 over 7 minus 1. 4 minus 2 is 2, 7 minus 1 is 6, and 2 over 6 reduces to 1 over 3. 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 meet 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, 10. And it's perpendicular, so the slope should be the same. The slope of the perpendicular bisector we said was negative 3, 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 3x 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.

Transcript for:Key Problems in Coordinate Geometry

Transcript for:
Key Problems in Coordinate Geometry