Honors Geometry Unit 13 Review: Coordinate Geometry

Analytic geometry uses equations to answer questions about where curves meet and how far apart things are. The most common problem is finding where a line and a circle intersect. For a nonvertical line, substitute its expression for $y$ into the circle equation and solve the quadratic. For a vertical line, substitute its constant value of $x$ and solve for $y$. The discriminant $b^2-4ac$ predicts the number of solutions: positive gives two points (secant), zero gives one point (tangent), and negative means the line misses the circle.

Worked example. Find where $y = x + 1$ meets $x^2 + y^2 = 25$. Substitute to get $x^2 + (x+1)^2 = 25$, so $2x^2 + 2x + 1 = 25$, which becomes $x^2 + x - 12 = 0$. Factor as $(x+4)(x-3)=0$, so $x = 3$ or $x = -4$. Plug back into the line to get the points $(3, 4)$ and $(-4, -3)$. Two points means the line is a secant.

A parabola is the set of points equidistant from a focus and a directrix, with the focus off the directrix. For a vertical parabola $y = a(x-h)^2 + k$, where $a \ne 0$, let $p = \frac{1}{4a}$. The focus is $(h, k+p)$ and the directrix is $y = k-p$; a negative $p$ puts the focus below the vertex. With focus $(0, 3)$ and directrix $y = -3$, the vertex is $(0,0)$, $p = 3$, and the equation is $y = \frac{1}{12}x^2$.

For a shortest path that must touch a line, reflect one endpoint across the line if the endpoints lie on the same side. The segment from the reflected point to the other endpoint gives the point where the path should touch the line. If the endpoints lie on opposite sides, the direct segment crosses the line and is shortest.

A coordinate proof places a figure on the plane, computes slopes, distances, or midpoints, and then states what those numbers prove. Smart placement saves algebra. For a general proof, put a vertex at the origin and a side along the x-axis, and use variables such as $(2a, 0)$ so midpoints come out without fractions.

For quadrilaterals, first decide what you need to show. A parallelogram needs both pairs of opposite sides parallel, both pairs congruent, or diagonals with the same midpoint. A rectangle is a parallelogram with one right angle or congruent diagonals. A rhombus is a parallelogram with perpendicular diagonals or four congruent sides. A square meets both the rectangle and rhombus conditions.

Worked example. Classify $A(0,0)$, $B(4,1)$, $C(5,5)$, $D(1,4)$. Each side has length $\sqrt{17}$ (for $AB$, $\sqrt{4^2+1^2}$), so it is a rhombus. Diagonal $AC$ has slope $1$ and $BD$ has slope $\frac{3}{-3} = -1$, so the diagonals are perpendicular, which agrees. Slopes of $AB$ and $BC$ are $\frac{1}{4}$ and $4$, whose product is $1$, not $-1$, so there is no right angle. The figure is a rhombus but not a square.

For triangles, use the distance formula for SSS congruence or SSS similarity. The centroid is the average of the three vertices.

Pick the line form that matches what you know. With a slope and a point, start in point-slope form $y - y_1 = m(x - x_1)$. With two points on a nonvertical line, find the slope first, then use point-slope. If the points have the same x-coordinate, the line is vertical and has equation $x = c$. Slope-intercept form $y = mx + b$ is useful for graphing nonvertical lines. Standard form is $Ax + By = C$, where $A$ and $B$ are not both zero; some classes ask for integer coefficients when possible and $A \ge 0$.

Nonvertical parallel lines share a slope, and two vertical lines are parallel. Perpendicular lines have slopes whose product is $-1$, so a line with slope $\frac{2}{3}$ is perpendicular to one with slope $-\frac{3}{2}$. Horizontal and vertical lines are perpendicular too, even though the reciprocal rule can't handle a slope of $0$.

A circle with center $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$. The signs inside flip, so $(x+1)^2 + (y-5)^2 = 7$ has center $(-1, 5)$ and radius $\sqrt{7}$.

Worked example. Convert $x^2 + y^2 - 2x + 6y + 1 = 0$ to standard form. Group and move the constant to get $(x^2 - 2x) + (y^2 + 6y) = -1$. Half of $-2$ is $-1$, squared is $1$. Half of $6$ is $3$, squared is $9$. Add both to each side to get $(x-1)^2 + (y+3)^2 = 9$. The center is $(1, -3)$ and the radius is $3$.

The distance formula finds the length of a segment: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. It is the Pythagorean theorem in disguise. When both changes are nonzero, the horizontal and vertical changes are the legs of a right triangle, and the segment is the hypotenuse. The formula also works when either change is zero. The order of the points doesn't matter because squaring removes negative signs.

The midpoint formula averages the coordinates: $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$. A quick sanity check is that each coordinate of the midpoint should land between the matching endpoint coordinates.

Worked example. For $A(-1, 2)$ and $B(5, 10)$, the distance is $\sqrt{(5-(-1))^2 + (10-2)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$. The midpoint is $\left(\frac{-1+5}{2}, \frac{2+10}{2}\right) = (2, 6)$. Each half of the segment should be $5$, and indeed $\sqrt{3^2 + 4^2} = 5$.

You can also work backward. If $M(2, 6)$ is the midpoint and one endpoint is $(-1, 2)$, the other endpoint is $(2 \cdot 2 - (-1), 2 \cdot 6 - 2) = (5, 10)$. Use distance when a question asks about length or congruence, and midpoint when it asks about a center or a bisector. On maps, multiply your distance by the scale at the end.