Regents Examination in Geometry: Geometry in the Coordinate Plane

Geometry in the Coordinate Plane

Length, Distance, and Midpoint

Key Ideas
Coordinate geometry takes the geometric building blocks of point and line and then maps them onto the coordinate plane by assigning each point a coordinate pair (X, Y).

Given the coordinates of the endpoints of a segment, (X1, Y1) and (X2, Y2), the following formulas allow us to calculate the length, midpoint, and point that divides a segment in a given ratio:

divide segment in a specified ratio:
,



Brief Review of the Coordinate Plane
Coordinate geometry associates every point with a coordinate pair (x, y), which locates the position of that point in the plane, as shown in Figure 7.1. The coordinates refer to the distance along a pair of perpendicular axes, the x-axis and the y-axis. The plane is divided into four quadrants (I, II, III, and IV) by the x-axis and y-axis.

The intersection of these axes is called the origin and is assigned the coordinates (0, 0).

The sign of each coordinate refers to a direction from the origin as follows:
positive x: right
negative x: left
positive y: up
negative y: down

The coordinate plane

The Distance Formula
The formula for calculating the distance between any two points is derived from the Pythagorean theorem. Consider two points, A and B, with coordinates A(x1, y1) and B(x2, y2) as shown in Figure. To calculate the length AB, we first complete a right triangle by sketching horizontal segment  and vertical segment . The lengths of these segments are (x2 − x1) and (y2 − y1). The Pythagorean theorem can now be used to find the distance, d, from A to B.

Using the Pythagorean theorem to calculate length

Distance Formula The distance, d, between two points with coordinate (x1, y1) and (x2, y2) is

Example 1
Find the length of the segments whose endpoints have coordinates (1, 2) and (5, 4).
Solution:



Math Fact
An alternative to remembering the distance formula is to complete the right triangle, count the lengths of the horizontal and vertical legs, and use the Pythagorean theorem.

Example 2
Are segments  and  congruent given the coordinates R(4, 1), S(10, −2), T(−5, 3), U(1, 6)?
Solution: Calculate the lengths of  and .

Congruent segments have equal lengths, therefore
.

Example 3
ΔABC has coordinates A(−4, −3), B(5, 0), and C(−3, 4). 

 is an altitude of the triangle, with D located at (−1, −2).

Solution: The area of a triangle equals . The height is the length of altitude , and the base is side AB. A sketch helps identify the appropriate side to use for the base.

Apply the distance formula:


Midpoint
The coordinates of the midpoint of a segment, (xMP, yMP), can be calculated from the coordinates of its endpoints, (x1, y1) and (x2, y2), using the midpoint formula.

Figureshows the midpoint of a segment and its coordinates.

Midpoint of a segment

Midpoint Formula

Example 1
Find the coordinates of the midpoint of , given coordinates P(1, 8) and Q(−5, 2).
Solution:

The coordinates of the midpoint are (−2, 5).
The midpoint formula can also be used in reverse to calculate the coordinates of an endpoint using the coordinates of the other endpoint and the midpoint.

Example 2
M is the midpoint of . Find the coordinates of point Q given P(3, 8) and M(7, −2).
Solution:


The coordinates of M are (11, −12).

Example 3
 is a diameter of circle P. The coordinates of D are (1, 4), and the coordinates of P are (5, 8). Find the coordinates of G.
Solution: The center of the circle, P, is the midpoint of the diameter. So we can use the midpoint formula to find the coordinates of G.


The coordinates of G are (9, 12).

Dividing a Directed Segment in a Specified Ratio
Besides dividing a segment into two congruent parts by finding a midpoint, we can divide a directed segment into two parts whose lengths are in any specified ratio. A directed segment is simply a segment in which we consider the first endpoint as the starting point and the second endpoint as the ending point.

To divide a directed segment in a specified ratio, we set up and solve the following proportion that relates the coordinates of the endpoints of the segment, P and Q, to the specified ratio.

The proportion is determined twice, once for the x-coordinate and once for the y-coordinate.

For example, consider directed segment  with coordinates J(1, −2) and K(11, 3). We want to find the coordinates of point L that divide it into a 2 : 3 ratio.
 is the smaller segment and  is the larger, as shown in Figure.

Directed line segment

To find the x-coordinate of point L, we use the proportion as follows:

Repeat for the y-coordinate:

The coordinates of L are (5, 0). Figure shows a graph of the segment and point L. If desired, a check can be performed by calculating the lengths of JL and LK and then showing they are in a 2 : 3 ratio.

Segment divided in a 2 : 3 ratio

Sometimes a point is stated to be a certain fraction of the way from one end of a segment to the other. For example, Q lies  of the way from A to B. In this case, the entire segment is broken up into 3 parts, with AQ equal to  of the total length and QB equal to the remaining . The ratio of the parts is therefore 1 : 2.

Example
Points U and V have coordinates U(−3, 7) and V(9, 1). Point W lies on segment  such that it is  of the way from U to V. Find the coordinates of W.
Solution: The segment is broken up into 6 parts, with UW equal to  the total length and WV equal to  of the total length. The ratio is therefore 1 : 5.

Repeat for the y-coordinate:

The coordinates of W are (−1, 6).

PERIMETER AND AREA USING COORDINATES

Key Ideas
The perimeter of any polygon can be calculated from the coordinates of the polygon’s vertices using the distance formula. To calculate the area, divide a polygon into rectangles and right triangles, whose areas can be easily calculated.

Perimeter
The perimeter of any polygon is easily calculated from the vertices. The length of each side is calculated using the distance formula and then all the sides are added. Alternatively, the figure can be graphed and each length can be calculated using the Pythagorean theorem. The distance formula and the Pythagorean theorem often result in radical expressions. Simplify radicals and combine like radicals when adding the lengths of the sides.

Example
Find the perimeter of the triangle with vertices A(–3, 4), B(5, 0), and C(3, 6).
Solution: Calculate the lengths of each of the three sides of ΔABC using the distance formula:



Area
The area of any polygon can be calculated by graphing the figure and sketching the smallest rectangle that bounds the polygon. The regions of the rectangle that go beyond the polygon are all right triangles or rectangles. Their areas can be subtracted from the area of the rectangle to give the area of the polygon.

Example 1
Find the area of the hexagon whose vertices are A(−4, 4), B(1, 3), C(4, −3), D(1, −6), E(−2, −5), and F(−5, −1).
Solution: The figure is graphed and bounded by the rectangle shown in the figure. The region outside the hexagon and within the rectangle is divided into the 8 regions shown.

 
Area of the bounding rectangle = 9 · 10 = 90

Area of region

Area of region

Area of region 3 = 3 · 1 = 3 
Area of region

Area of region

Area of region

Area of region 7 = 1 · 3 = 3
Area of region

Area of hexagon = 90 − (2.5 + 2.5 + 3 + 9 + 4.5 + 1.5 + 3 + 6) = 58

Calculating the area of a figure with a curved perimeter can be estimated using the reverse of the above technique.

The curved figure is filled in with squares, rectangles, and triangles as close as desired to the figure’s perimeter.

Smaller squares, rectangles, and triangles allow you to fill in closer to the perimeter, giving a more accurate estimate.

Example 2
Estimate the area of the ellipse shown in the accompanying figure.

 
Solution: We can estimate the area by filling in the ellipse with 5 rectangles. Triangles could have been used as well to get a more accurate estimate of the area.

 
Calculate the areas of the rectangles.
Rectangle 1 area = 7 · 6 = 42
Rectangle 2 area = 4 · 2 = 8
Rectangle 3 area = 5 · 1 = 5
Rectangle 4 area = 4 · 2 = 8
Rectangle 5 area = 5 · 1 = 5
The approximate area of the ellipse is 42 + 8 + 5 + 8 + 5 = 68.
For comparison, the actual area of the ellipse is 69. So our estimate was within 2% of the actual area.

SLOPE AND EQUATIONS OF LINES

Slope
The slope of a straight line is a measure of its steepness, as shown in Figure 7.6. It is also the change in the y-coordinate divided by the change in the x-coordinate between any two points on the line. The change in the y-coordinate is called the rise, and the change in the x-coordinate is called the run. Slope is calculated from the coordinates of each point using the slope formula.

Slope of a line

Slope Formula:

In Figure, the slope is calculated between points A and B. The rise is the difference in the y-coordinates: 2 − (−1) = 3. The run is the differences in the x-coordinates: 0 − (−2) = 2. The result is a slope of . The rise and run can be counted from a graph if you can identify 2 points on the line whose coordinates are easily determined.

Keep in mind several important features of slope:

1. The slopes between any pair of points on a straight line are equal.
2. The slope between two points A and B on a straight line will be the same whether it is calculated from A to B or B to A.
3. Slope is positive if the straight line is directed up and to the right.
4. Slope is negative if the straight line is directed down and to the left.
5. The slope of any horizontal line equals zero since the rise is zero.
6. The slope of any vertical line is undefined since the run is zero and we cannot divide by zero.
7. The slope of parallel lines are equal.
8. The slope of perpendicular lines are negative reciprocals (except for horizontal and vertical lines, which are also perpendicular).

Example
Find the slope of each line in the accompanying figure.

Solution: For each line, determine the coordinates of two points. Then apply
.

with R(−2, 3) and S(2, −1).


with M(0, 3) and N(5, 5).


with

with Q(4, 1) and P(4, −2). 

is undefined

Collinearity
Slope can be used to determine the collinearity of three points. Given points A, B, and C, the three points will be collinear if the slope of
 equals the slope of .

Example
Are points D(1, 4), E(4, 6), and F(7, 8) collinear?
Solution: Using
, the slope of  and the slope of . The slopes are equal. So the points are collinear.

Equations of Lines
A straight line in the x-y-plane is described by a function that relates a value of the independent variable x to a value of the dependent variable y. The functions that describe straight lines are linear. A linear function is one in which the variables are raised only to the first power.

Two common ways to write the equation of a line are the point-slope form and the slope-intercept form.

Point-Slope Form of a Line: where m is the slope and where x1 and y1 are the coordinates of any point on the line. Slope-Intercept Form of a Line: where m is the slope and b is the y-intercept of the line. The y-intercept is the y-coordinate at which the line crosses the y-axis.

Figure shows the graph of the line whose point–slope equation is y − 3 = 4(x − 1). This represents a line that passes through the point (1, 3) and has a slope of 4. The line can be graphed using (1, 3) as a starting point and then graphing additional points by following a rise/run of 4. The same line in slope-intercept form would be y = 4x − 1. The slope is 4, and the y-intercept is −1. Start at the point (0, −1) and follow a rise/run of 4 to locate additional points on the line.

Graph of the line y − 3 = 4( x − 1)


Math Fact
The x-intercept of a line occurs when the y-coordinate is zero, and the y-intercept occurs when the x-coordinate is zero. Either intercept can be found from the equation of the line by setting the other coordinate equal to zero.


Example 1
Graph the line  and the line .
Solution: The line  has a slope of  and a y-intercept of −3. Graph a point at (0, −3), and then continue the line with a slope of . The line  passes through the point (1, 1) and continues with a slope of . The graphs are shown in the accompanying figure.


Example 2
Write the equations of the lines  and  shown in the accompanying figure.

 
Solution: Using points G and Q, the slope of  is , and the equation of  in point-slope form is .
Using points W and F, the slope of . The y-intercept is −3. The slope-intercept form of the equation of
 is y = −3x − 3.

When written in slope-intercept form, the equation allows you to calculate the corresponding y-coordinate for any desired x-coordinate easily. For example, consider the line given by the equation y = 2x + 3. When x = 1, y = 2(1) + 3 = 5. When x = 2, y = 2(2) + 3 = 7. The resulting coordinate pairs are (1, 5) and (2, 7). These points lie on the line. Think about it the other way. Every point that lies on the line must satisfy the equation of the line. This means if we substitute the x- and y-values from the coordinate pair (x, y), the equation will be balanced.

Example 3
Do the points (4, 8) and (6, 9) lie on the line whose equation is y = 3x − 4?
Solution: By substituting x = 4 and y = 8 into y = 3x − 8, we get 8 = 3(4) − 4 = 8. The equation is balanced, and (4, 8) is on the line.
By substituting x = 6 and y = 9 into y = 3x − 8, we get 9 = 3(6) − 4 ≠ 14. The equation is not balanced, and (6, 9) is not on the line.

Writing the Equation of a Line Given Two Points or a Slope and Point
When given two points, the equation of the line passing through those points can be written by first calculating the slope and then substituting the slope and the coordinates of one point into the point-slope form of the line.

Math Fact
If you are given two points, the equation can be found by graphing the two points and drawing the line between them with a straightedge. You can determine the slope by counting the rise and the run from the graph. You can find the y-intercept by following the line back to the y-axis. You now have the slope and y-intercept to use in the equation y = mx + b.


Example
Write the equation of the line passing through (1, 5) and (3, 11). Give both the point-slope form and the slope-intercept form.
Solution:
. Substituting the slope and the first coordinate pair into y − y1 = m(x − x1) yields the point-slope equation y − 5 = 3(x − 1).
Find the slope-intercept form by solving for y:

EQUATIONS OF PARALLEL AND PERPENDICULAR LINES

Key Ideas
We have two theorems regarding slopes of parallel and perpendicular lines:
If two lines are parallel, then their slopes are equal.
If two lines are perpendicular, then their slopes are negative reciprocals.
These can be used to write equations of a line parallel or perpendicular to a given line or to write the equation of the perpendicular bisector of a given segment.

Slopes and Equations of Parallel and Perpendicular Lines

When given a line, there are an infinite number of parallel lines. All have the same slope but different y-intercepts. Figure 7.8 shows a family of parallel lines and their equations. Notice that only the y-intercept changes.

Parallel lines

You can determine if two lines are parallel by comparing their slopes. If the equations are not in either slope-intercept or point–slope forms, rewrite them in one of those forms so the slope can be easily identified.

Example 1
Are the lines y = 4x + 2 and 2y − 8x − 10 = 0 parallel?
Solution: The slope of the first equation is 4. The second equation must be rearranged:

The second line also has a slope of 4, so the lines are parallel.

Example 2
What is the slope of a line parallel to 6x + 3y = 12?
Solution: Rearrange the equation so it is in slope-intercept form:

The slope of the line is −2, so a parallel line would also have a slope of −2.

The negative reciprocal of a slope m is = . There are an infinite number of lines perpendicular to a given line. They will all have the same slope but different y-intercepts. Figure 7.9 shows a family of perpendicular lines. Since all the lines perpendicular to 
have the same slope, they must be parallel to each other.

Perpendicular lines

The product of a pair of negative reciprocals is always −1. If two lines are perpendicular, the product of their slopes is −1.

Example 1
What is the slope of a line perpendicular to x + 5y = 10?
Solution: Rearrange into slope-intercept form:

The slope of the line is . The slope of a perpendicular line is the negative reciprocal, which is 5.

We can combine what we have learned about parallel lines, perpendicular lines, and equations of lines to write the equation of a line parallel or perpendicular to a given line and through a given point. The given line is used to find the desired slope. The given point becomes the x1 and y1 in the point-slope equation of the line. Remember that the y-intercept of the new equation may be different from the y-intercept of the given line.

Example 2
Find the equation of the line parallel to y = 3x − 7 and passing through the point (4, 1).
Solution: The slope of the given line is 3. By substituting m = 3, x1 = 4, and y1 = 1 into the point-slope equation, we get:


Example 3
Find the equation of the line perpendicular to y = 6x + 3 and passing through the point (3, 5).
Solution: The slope of the given line is 6, so the slope of the perpendicular line is . By substituting , x1 = 3, and y1 = 5 into the point-slope equation, we get:


Equation of a Perpendicular Bisector
Using the tools above, we can find the equation of a perpendicular bisector of a segment. Recall that the bisector of a segment passes through the segment’s midpoint and that we have a formula for finding the midpoint of a segment given the two endpoints. First calculate the coordinates of the midpoint and the slope of the segment. Then substitute the negative reciprocal of the slope and the coordinates of the midpoint into the point-slope equation.

Figure shows the line y = −x, which is the perpendicular bisector of segment . The line y = −x passes through the midpoint M of . The slope of  is 1, and the slope of y = −x is −1, which are negative reciprocals.

Perpendicular bisector of a segment


Math Fact
You must find the midpoint of the segment when asked for the equation of a perpendicular bisector. What would happen if you used one of the endpoints? You would get a perpendicular line, but it would pass through the endpoint instead of the midpoint. In the Figure, the line would pass through A or B instead of M.

Example
Find the equation of the perpendicular bisector of the segment with endpoints A(−2, 1) and B(4, 3).
Solution: First find the midpoint:

The midpoint is (1, 2).
Find the slope:

Use the negative reciprocal of the slope, which is −3.
Now substitute into the point slope equation:

EQUATIONS OF LINES AND TRANSFORMATIONS

Finding the Line of Reflection
We know that a line of reflection is always the perpendicular bisector of the segment joining corresponding points. Therefore, when given a pair of corresponding points, you can find the line of reflection using the procedure to find the perpendicular bisector.

Example
The reflection of T(4, 1) over the line  is T′(5, 4). Write the equation of .
Solution: First find the slope of :

The slope of the line of reflection is the negative reciprocal.
Slope of 
Next find the midpoint of :

By using the midpoint and slope, we get:


Finding the Image of a Point After a Reflection
Earlier, we saw rules for finding the image of a point after a reflection over horizontal and vertical lines and over the lines y = x and y = −x. For any other lines, the best approach is graphical. Find the slope of the given line of reflection. Then use the negative reciprocal of the slope to trace a path from the preimage to the line of reflection. Continue the same distance on the other side to locate the image.

Example
Point A(−1, 2) is reflected over the line y = 3x − 5 to A′. Find the coordinates of A′.
Solution: First graph A and the given line. The slope of the line is 3, and the negative reciprocal is . From A, follow a path with a slope of
 to the line, which is 3 jumps right and 1 jump down in this case. Continue the same number of jumps on the other side of the line to (5, 0). The coordinates of A′ are (5, 0).


Finding the Equation of a Line After a Dilation
When given a line, the equation of the image of the line after a dilation through the origin can be found as follows:
Find the coordinates of the y-intercept, and dilate by the specified dilation.
The new line has the same slope as the original since slope is preserved.

Example
Write an equation for the line that represents the dilation of 2y − x + 4 = 0 by a scale factor of 3 through the origin.
Solution: Rewrite the equation in y = mx + b form

The slope of the preimage is , and the y-intercept has coordinates (0, −2).
Dilate the y-intercept by a factor of 3 to get the new intercept (0, −6), and use the same slope.
The equation of the image is .

THE CIRCLE

Key Ideas
The center-radius equation of the circle is (x − h)2 + (y − k)2 = r2, where (h, k) is the center of the circle and r is the radius. In general form, the equation is x2 + Cx + y2 + Dy + E = 0. We can convert from general form to center-radius form by completing the square.

Equation of the Circle—Center-Radius Form
Circles are one of the conic sections, along with parabolas, ellipses, and hyperbolas. The name conic section comes from the fact that each curve is obtained by taking a cross-sectional slice from a cone. A circle can be represented by an algebraic relationship between the x- and y-coordinates.
The equation is derived from the Pythagorean theorem. Figure shows circle C with radius r and C located at coordinates (h, k). Let point P be any point on the circle with coordinates (x, y).

Recall the theorem:

Circle in the coordinate plane

The set of points a distance r from a given point P is a circle centered at P with radius r.
No matter where on the circle we choose point P, the distance PC must be equal to r. If we complete the right triangle with r as the hypotenuse, the base has length (x − h) and the height has length (y − k).

The Pythagorean theorem or the distance formula applied to the triangle gives:

Graphing a circle from the equation in (x − h)2 + (y − k)2 = r2 form is very straightforward. The center (h, k) and the radius r are easily identified. Remember that the constant term on the right is the square of the radius—we need to take the square root to find the radius. Also note that the h- and k-terms are subtracted, so be careful with the signs. Once the center and radius are determined, graph a point at the center. The first four points on the circle are plotted by translating the center up r units, down r units, right r units, and left r units. Finally complete the circle passing through the first four points. A careful graph can be made using a compass, or a rough sketch can be made by drawing the circle freehand.


Math Fact
A circle is not a function, and it is not one-to-one. For every x-coordinate, there are two corresponding y-coordinates. Also there are two different x-coordinates that lead to the same y-coordinate.


Example 1
Graph the equation (x − 2)2 + (y + 1)2 = 9.
Solution: The center is located at (2, −1). Watch out for the signs! In the equation, 2 is subtracted from x and −1 is subtracted from y. The radius is . We plot the center point at (2, −1) and then points up, down, right, and left 3 from the center. Then complete the circle.


Example 2
Write the equations of the three circles shown in the accompanying figure.

 
Solution:
The center of circle A is (−1, 3), and the radius is 2. So the equation is (x + 1)2 + (y − 3)2 = 4.
The center of circle B is (3, 1), and the radius is 1. So the equation is (x − 3)2 + (y − 1)2 = 1.
The center of circle C is (0, −1), and the radius is 6. So the equation is x2 + (y + 1)2 = 36.
If the coordinates of the center and one point on the circle are given, we can apply the distance formula to find the radius. If only the two endpoints of a diameter are given, the midpoint and distance formulas can be used to find the missing center and radius.

Example 3
Find the equation of the circle with center C at (2, 7) and point P(4, 12) on the circle.
Solution: The distance formula used on the center C and point P will give the radius, r, of the circle:

The radius squared is 29. Substituting into (x − h)2 + (y − k)2 = r2 yields (x − 2)2 + (y − 7)2 = 29.

Example 4
 is a diameter of circle O. The coordinates of B are (−4, 9), and the coordinates of D are (6, 3). Find the equation of the circle.
Solution: The first step is to find the center of the circle using the midpoint formula:

The center O has coordinates (1, 6).
Now we can find the radius by calculating the distance from point O to either B or D. It doesn’t matter which one we pick since any radius has the same length.

When using D, we get:

By substituting the coordinates of O and the radius into the equation of the circle, we get:

The equation of a circle can also be used to determine if a particular point lies on the circle. As with equations describing lines, parabolas, and other curves, the equation of a circle will be satisfied by the coordinates of any point that lies on the circle.

Example 5
Does the point (−9, 5) lie on the circle described by (x + 4)2 + (y − 1)2 = 41?
Solution: By substituting the coordinates into the equation, we find:

The equation is balanced, so the point does lie on the circle.


General Form of the Equation of the Circle
We can expand the two squared terms in the center-radius form of the circle.

General Form of the Equation of a Circle

For example, the circle (x + 2)2 + (y + 5)2 = 4 in general form is:

Going from general form to center-radius form involves completing the square. First rearrange the equation so all the x-terms and all the y-terms are grouped on one side and the constant is on the other.

Using the previous equation as an example, we get:

)))2 or 25.

After adding the required constants on both sides, we can factor the terms with x’s and the terms with y’s to get back where we started from.


Example
Find the coordinates of the center and the length of the radius of a circle whose equation is x2 + y2 + 6x + 8y + 12 = 0.
Solution: Using the procedure for completing the square:

x2 + y2 + 6x + 8y + 12 = 0           x2 + 6x + y2 + 8y = −12 x2 + 6x + 9 + y2 + 8y + 16 = −12 + 9 + 16 (x + 3)2 + (y + 4)2 = 13

group the x and y terms calculate the required constants add the constants to both sides complete the square