Coordinate Geometry Coordinate geometry involves graphs in the (x, y) coordinate plane. For the SAT and the ACT, you should be especially proficient with the coordinate geometry of linear functions. You should also know the basics about the graphs of quadratics. On the ACT, you should also know the basics of other conic sections: circles, ellipses, and hyperbolas.

Linear Functions A Linear Function forms a straight line when graphed in the coordinate plane. Although linear functions can be written in several forms, slope intercept form is the most common and the most useful. Slope intercept form: y = mx + b, where m = slope and b = y-intercept Slope = rise/run = (change in y)/(change in x) = (y2 – y1)/(x2 – x1) The slope of a vertical line is undefined. The slope of a horizontal line is zero. Parallel lines have the same slope. Perpendicular lines have negative reciprocal slopes. For instance, a line with a slope of 3 would be perpendicular to a line with a slope of -1/3, and a line with a slope of -4/7 would be perpendicular to a line with a slope of 7/4. y-intercept = The y-coordinate at which the line intercepts the y-axis. To find the y-intercept, put the equation in slope intercept form; the y-intercept is equal to b. Alternatively, you can find the y-intercept by substituting 0 for x in the equation and solving for y. x-intercept = The x-coordinate at which the line intercepts the x-axis. To find the x-intercept, substitute 0 for y in the equation and solve for x. 1) What is the equation of a line that contains the points (2, 4) and (-2, 6)? Solve this problem by using the slope formula to find the slope. (6 – 4)/(-2 – 2) = 2/-4 = -1/2. Plug this slope in for m in y = mx + b, and plug in the x and y coordinates from one of the two points into the equation for x and y, respectively: 4 = (-1/2)(2) + b Solve algebraically for b. 4 = -1 + b b = 5. Now that you know both m and b, you can write the equation of the line in yintercept form: y = -1/2x + 5. 2) What is the equation of a line that is perpendicular to 4y = 16x – 5 and passes through the point (1, -4)? This problem is almost identical to problem 1. The only difference is in how you find the slope. First, you must find the slope of 4y = 16x – 5. To find the slope of this line, put it into yintercept form and the slope will be equal to m. y = 4x – 5/4, so the slope of this line is 4. Because the line you are trying to find is perpendicular to this line, its slope must be the negative reciprocal of 4, which is -1/4. Plug in this slope for m in y=mx + b, and plug in the x and

y coordinates of the point (1, -4) for x and y, respectively: -4 = (-1/4)(1) + b. Solve this equation for b, the y-intercept. b = -15/4. Now that you know both m and b, you can write the equation of the line in y-intercept form: y = -1/4x – 15/4. Midpoint of a Line Segment: ((x1+x2)/2, (y1+y2)/2), where (x1, y1) and (x2, y2) are the two endpoints. The above formula is important, but you can probably avoid memorizing it if you understand it conceptually. To find the midpoint of a line segment, all you need to do is average the xcoordinates of the two endpoints to get the x-coordinate of the midpoint and average the ycoordinates of the two endpoints to get the y-coordinate of the midpoint. 3) What is the midpoint of the line segment with endpoints (1, -3) and (-7, -1)? To solve, use the midpoint formula, or simply average the x’s to get the x-coordinate of the midpoint and average the y’s to get the y-coordinate of the midpoint. Either way, you get (1 + 7)/2 = -3 for the x-coordinate and (-3 + -1)/2 = -2 for the y-coordinate. Therefore, the midpoint is (-3, -2). 4) One endpoint of a line segment is (3, 8) and the midpoint is (-1, 15). What is the other endpoint? Be careful on this problem, since it gives you one endpoint and the midpoint and asks for the other endpoint. It would be easy to mistakenly find the midpoint of the two points given. One way to find the other endpoint is to use the formula. (3 + x)/2 = -1 and (8 + y)/2 = 15. Solving these equations, you get x = -5 and y = 22, so the other endpoint is (-5, 22). You could also do this problem without using the equation just by realizing that the midpoint is halfway between both endpoints. Therefore, because 3 is 4 more than -1, the x-coordinate of the other endpoint must be 4 less than -1, so it must be 5. Because 8 is 7 less than 15, the y-coordinate of the other endpoint must be 7 more than 15, so it must be 22. Distance formula: The distance between the points (x1, y1) and (x2, y2) can be calculated using the following formula. D=√ However, it is not necessary to memorize this formula if you can understand it conceptually. This formula is simply the Pythagorean Theorem, . To find the distance between two points in the coordinate plane, conceptualize a right triangle. If you understand this concept, you should not need to actually draw the triangle. One leg of this triangle is the difference of the x-coordinates of the two points. The other leg of the triangle is the difference of the y-coordinates of the two points. Find these two differences, and plug them into the Pythagorean Theorem for a and b. The distance between the two points is equal to the hypotenuse of the triangle you have conceptualized, so use the Pythagorean Theorem to solve for c, the distance between the two points. 5) What is the distance between the points (4,-3) and (9, 5)?

To solve, you could plug into the distance formula: √ =√ = 9.434. Alternately, you could conceptualize it as a triangle. One leg is the difference of √ the x’s, the other leg is the difference of the y’s, and the hypotenuse is the distance between the two points. Hence, 52 + 82 = d2 d = √ 9.434.

Conic Sections Parabolas, hyperbolas, circles, and ellipses are known as conic sections. The SAT only tests students on parabolas, while the ACT requires students to have some familiarity with all conic sections. Parabolas Standard form: y = ax2 + bx + c The most common conic section tested on the ACT, and the only conic section tested on the SAT, is the parabola, which is the graph of a quadratic equation. Quadratics and parabolas are not discussed in depth in this discussion of conic sections. See quadratics for more information. For the purposes of this discussion, just recognize that any equation that contains an x 2 but no y2 is a parabola and that an equation that contains an x2 and a y2 can be a circle, an ellipse, or a hyperbola.

The conic sections below, Circles, Ellipses, and Hyperbolas are tested on the ACT only. Unless you are trying to get a 32 or higher on the ACT math, it is probably not worth worrying about ellipses or hyperbolas. All ACT takers, however, should be comfortable with circles in the coordinate plane. Circles Standard form for a circle: (x – h)2 + (y – k)2 = r2 (h, k) = center point r = radius

Other than parabolas, circles are the conic section that appears most frequently on the ACT. They appear frequently enough that you should definitely memorize the equation above; most circle problems are very easy if you just know this equation, as they involve little more than plugging in values for the center and radius. Note that the equation for a circle is essentially

the Pythagorean Theorem. Indeed, a circle that is centered at the origin (0, 0) has the equation x2 + y2 = r2. It is important to be able to recognize a circle based on its equation even if it is not in standard form. If an equation contains both an x2 and a y2, and they contain the same coefficients and the same sign (positive or negative) while on the same side of the equation, the graph is a circle. For example, x2 + y2 = 16 and 3x2 + 3y2 = 25 are both circles. Ellipses Standard form for an ellipse:

–

–

+

=1

An ellipse is essentially a circle that has been stretched either vertically or horizontally into what we commonly refer to as an “oval.” Imagine if you were to divide both sides of the standard form for a circle by r2. This algebraic manipulation would leave you with

–

+

–

= 1 as the equation of the circle. Comparing this to the equation of an ellipse, you can see that an ellipse is essentially a circle with an “x radius” that is different from the “y radius.” Technically, it is mathematically improper to refer to a and b as the x radius and y radius, respectively, but it is a useful way of thinking about ellipses for the purposes of the ACT. The only other terminology you should know for ellipses is that the major axis is equal to twice the larger of the two “radii” and the minor axis is equal to twice the smaller of the two “radii.” It is important to be able to recognize an ellipse based on its equation even if it is not in standard form. If an equation contains both an x2 and a y2, and these terms have different coefficients but the same sign (positive or negative) while on the same side of the equation, the graph is an ellipse. For example, 2x2 + y2 = 16 and 3x2 + 8y2 = 15 are both ellipses. Hyperbolas Standard form for a hyperbola:

–

–

–

=1

Hyperbolas come up only on rare occasion on the ACT. The only thing you will likely have to do regarding hyperbolas is be able to recognize one based on the equation, even if it is not in

standard form. If an equation contains both an x2 and a y2, and these terms have opposite signs (one is positive and one is negative) while on the same side of the equation, the graph is a hyperbola. For example, x2 – y2 = 9 and -2x2 + 5y2 = 12 are both hyperbolas. 6) What is the equation of a circle with center point (3, -4) and a radius of 8? Plugging r, h, and k into the proper places in the equation for standard form of a circle: (x – 3)2 + (y + 4)2 = 64 7) The graph of the equation 2x2 + 3y2 = 8 would be which of the following? The graph of this equation is an ellipse. To determine this, you do not need to put this equation into standard form; you only need to recognize that any equation that contains an x 2 and a y2 with different coefficients but the same sign (positive or negative) while on the same side of the equation will form an ellipse when graphed.