Plane Coordinate Geometry and Parametric Equations Notes

The Coordinate Plane

  • Graphing regions in the coordinate plane involves describing and sketching sets of points denoted by specific conditions on their coordinates (x,y)(x, y).

  • Example 1: Describing various regions:

    • (a) The set {(x,y)∣x≥0}\{(x, y) \mid x \ge 0\} represents all points in the coordinate plane where the xx-coordinate is non-negative. This corresponds to the right half-plane, including the yy-axis.

    • (b) The set {(x,y)∣y=1}\{(x, y) \mid y = 1\} represents all points where the yy-coordinate is exactly 11, resulting in a horizontal line passing through (0,1)(0, 1).

    • (c) The set \{(x, y) \mid |y| < 1\} represents the region between the horizontal lines y=−1y = -1 and y=1y = 1, not including the lines themselves.

The Distance and Midpoint Formulas

  • Distance Formula

    • The distance d(A,B)d(A, B) between any two points A(x1,y1)A(x_1, y_1) and B(x2,y2)B(x_2, y_2) in the Cartesian plane is calculated using the formula:     d(A,B)=(x2−x1)2+(y2−y1)2d(A, B) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

  • Example 2: Determining proximity to a point:

    • To find which point, P(1,−2)P(1, -2) or Q(8,9)Q(8, 9), is closer to A(5,3)A(5, 3), distances are calculated separately:

      • d(P,A)=(5−1)2+[3−(−2)]2=42+52=16+25=41d(P, A) = \sqrt{(5 - 1)^2 + [3 - (-2)]^2} = \sqrt{4^2 + 5^2} = \sqrt{16 + 25} = \sqrt{41}

      • d(Q,A)=(5−8)2+(3−9)2=(−3)2+(−6)2=9+36=45d(Q, A) = \sqrt{(5 - 8)^2 + (3 - 9)^2} = \sqrt{(-3)^2 + (-6)^2} = \sqrt{9 + 36} = \sqrt{45}

    • Since d(P, A) < d(Q, A) (as \sqrt{41} < \sqrt{45}), point PP is closer to point AA.

  • Midpoint Formula

    • The midpoint of a line segment connecting points A(x1,y1)A(x_1, y_1) and B(x2,y2)B(x_2, y_2) is found by averaging the coordinates:     (x1+x22,y1+y22)(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})

  • Example 3: Proving a quadrilateral is a parallelogram:

    • Given vertices P(1,2)P(1, 2), Q(4,4)Q(4, 4), R(5,9)R(5, 9), and S(2,7)S(2, 7), a quadrilateral is a parallelogram if its diagonals bisect each other (i.e., they share the same midpoint).

    • Midpoint of diagonal PRPR: (1+52,2+92)=(3,112)(\frac{1 + 5}{2}, \frac{2 + 9}{2}) = (3, \frac{11}{2})

    • Midpoint of diagonal QSQS: (4+22,4+72)=(3,112)(\frac{4 + 2}{2}, \frac{4 + 7}{2}) = (3, \frac{11}{2})

    • Since both diagonals share the midpoint (3,5.5)(3, 5.5), they bisect each other, proving PQRSPQRS is a parallelogram.

Equations of Lines

  • Fundamental Principle of Analytic Geometry

    • A point (x,y)(x, y) lies on the graph of an equation if and only if its coordinates satisfy that equation.

  • Slope of a Line

    • The slope mm of a nonvertical line passing through A(x1,y1)A(x_1, y_1) and B(x2,y2)B(x_2, y_2) is defined as the ratio of the rise to the run:     m=riserun=y2−y1x2−x1m = \frac{\text{rise}}{\text{run}} = \frac{y_2 - y_1}{x_2 - x_1}

    • The slope of a vertical line is undefined.

    • Slope is independent of the choice of points on the line, which can be verified using similar triangles.

  • Example 4: Finding slope through two points:

    • For points P(2,1)P(2, 1) and Q(8,5)Q(8, 5), the slope is calculated as:     m=5−18−2=46=23m = \frac{5 - 1}{8 - 2} = \frac{4}{6} = \frac{2}{3}

  • Point-Slope Form

    • The equation of a line passing through (x1,y1)(x_1, y_1) with slope mm is:     y−y1=m(x−x1)y - y_1 = m(x - x_1)

  • Example 5: Finding an equation with a point and slope:

    • (a) Line through (1,−3)(1, -3) with slope m=−12m = -\frac{1}{2}:

      • Using y−(−3)=−12(x−1)y - (-3) = -\frac{1}{2}(x - 1)

      • y+3=−12(x−1)y + 3 = -\frac{1}{2}(x - 1)

      • Multiplying by 22: 2y+6=−x+12y + 6 = -x + 1

      • Rearranging into general form: x+2y+5=0x + 2y + 5 = 0

    • (b) Linear Sketching: A slope of −12-\frac{1}{2} indicates that for every 22 units moved to the right, the line drops by 11 unit.

  • Example 5 (Continued): Line through two given points:

    • For points (−1,2)(-1, 2) and (3,−4)(3, -4), the slope is:     m=−4−23−(−1)=−64=−32m = \frac{-4 - 2}{3 - (-1)} = \frac{-6}{4} = -\frac{3}{2}

    • Using point-slope form with (−1,2)(-1, 2):

      • y−2=−32(x+1)y - 2 = -\frac{3}{2}(x + 1)

      • 2y−4=−3x−32y - 4 = -3x - 3

      • Rearranging: 3x+2y−1=03x + 2y - 1 = 0

  • Slope-Intercept Form

    • An equation of a line with slope mm and yy-intercept bb is simplified from y−b=m(x−0)y - b = m(x - 0) to:     y=mx+by = mx + b

  • Example 6: Evaluating slope-intercept forms:

    • (a) Finding an equation with m=3m = 3 and b=−2b = -2 yields y=3x−2y = 3x - 2.

    • (b) To find the slope and intercept of 3y−2x=13y - 2x = 1:

      • Isolate yy: 3y=2x+13y = 2x + 1

      • Divide by 33: y=23x+13y = \frac{2}{3}x + \frac{1}{3}

      • Slope m=23m = \frac{2}{3}, yy-intercept b=13b = \frac{1}{3}.

Parallel and Perpendicular Lines

  • Parallel Lines

    • Two nonvertical lines are parallel if and only if they have identical slopes (m1=m2m_1 = m_2).

  • Example 7: Finding a parallel line:

    • Find a line through (5,2)(5, 2) parallel to 4x+6y+5=04x + 6y + 5 = 0.

    • First, determine the slope of the given line: 6y=−4x−5  ⟹  y=−23x−566y = -4x - 5 \implies y = -\frac{2}{3}x - \frac{5}{6}. Thus, m=−23m = -\frac{2}{3}.

    • The parallel line also has m=−23m = -\frac{2}{3}. Using point-slope form:

      • y−2=−23(x−5)y - 2 = -\frac{2}{3}(x - 5)

      • 3y−6=−2x+103y - 6 = -2x + 10

      • Resulting equation: 2x+3y−16=02x + 3y - 16 = 0

  • Perpendicular Lines

    • Two lines with slopes m1m_1 and m2m_2 are perpendicular if and only if m1m2=−1m_1 m_2 = -1, meaning their slopes are negative reciprocals: m2=−1m1m_2 = -\frac{1}{m_1}.

    • Horizontal lines (slope 00) are always perpendicular to vertical lines (undefined slope).

  • Example 8: Finding a perpendicular line:

    • Find a line perpendicular to 4x+6y+5=04x + 6y + 5 = 0 that passes through the origin (0,0)(0, 0).

    • The slope of the given line is −23-\frac{2}{3}, so the perpendicular slope is 32\frac{3}{2}.

    • Using point-slope form: y−0=32(x−0)y - 0 = \frac{3}{2}(x - 0), which simplifies to y=32xy = \frac{3}{2}x.

Distance Between a Point and a Line

  • The distance dd from a point Q(x0,y0)Q(x_0, y_0) to the line defined by ax+by+c=0ax + by + c = 0 is given by the formula:     d=∣a(x0)+b(y0)+c∣a2+b2d = \frac{|a(x_0) + b(y_0) + c|}{\sqrt{a^2 + b^2}}

  • Example 9: Distance calculation:

    • For line 2x+4y−5=02x + 4y - 5 = 0 and point Q(−3,2)Q(-3, 2), the distance is:     d=∣2(−3)+4(2)−5∣22+42=∣−6+8−5∣4+16=∣−3∣20=325d = \frac{|2(-3) + 4(2) - 5|}{\sqrt{2^2 + 4^2}} = \frac{|-6 + 8 - 5|}{\sqrt{4 + 16}} = \frac{|-3|}{\sqrt{20}} = \frac{3}{2\sqrt{5}}

  • Example 10: Distance from the origin:

    • If the point is at the origin (0,0)(0, 0), the formula simplifies to:     d=∣c∣a2+b2d = \frac{|c|}{\sqrt{a^2 + b^2}}

    • For the line 3x+4y−6=03x + 4y - 6 = 0:     d=∣−6∣32+42=69+16=65d = \frac{|-6|}{\sqrt{3^2 + 4^2}} = \frac{6}{\sqrt{9 + 16}} = \frac{6}{5}

Circles

  • Equation of a Circle

    • The standard form for an equation of a circle with center (h,k)(h, k) and radius rr is:     (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2

    • If the circle is centered at the origin (0,0)(0, 0), the equation simplifies to:     x2+y2=r2x^2 + y^2 = r^2

  • Example 11: Graphing circles:

    • (a) x2+y2=25x^2 + y^2 = 25 is a circle centered at (0,0)(0, 0) with radius r=5r = 5 (since 52=255^2 = 25).

    • (b) (x−2)2+(y+1)2=25(x - 2)^2 + (y + 1)^2 = 25 is a circle centered at (2,−1)(2, -1) with radius r=5r = 5.

  • Example 12: Determining circle equations:

    • (a) Given radius r=3r = 3 and center (2,−5)(2, -5), the equation is:     (x−2)2+(y+5)2=9(x - 2)^2 + (y + 5)^2 = 9

    • (b) Given diameter endpoints P(1,8)P(1, 8) and Q(5,−6)Q(5, -6), find the center and radius:

      • Center (midpoint of PQPQ): (1+52,8−62)=(3,1)(\frac{1 + 5}{2}, \frac{8 - 6}{2}) = (3, 1)

      • Calculate r2r^2 (distance from center to PP): r2=(3−1)2+(1−8)2=22+(−7)2=4+49=53r^2 = (3 - 1)^2 + (1 - 8)^2 = 2^2 + (-7)^2 = 4 + 49 = 53

      • Final equation: (x−3)2+(y−1)2=53(x - 3)^2 + (y - 1)^2 = 53

    • Expanded form of the above equation:

      • (x−3)2+(y−1)2=53(x - 3)^2 + (y - 1)^2 = 53

      • x2−6x+9+y2−2y+1=53x^2 - 6x + 9 + y^2 - 2y + 1 = 53

      • x2−6x+y2−2y=43x^2 - 6x + y^2 - 2y = 43

  • Example 13: Identifying a circle from a general equation:

    • To show x2+y2+2x−6y+7=0x^2 + y^2 + 2x - 6y + 7 = 0 represents a circle, complete the square for xx and yy:

      • Rearrange: (x2+2x)+(y2−6y)=−7(x^2 + 2x) + (y^2 - 6y) = -7

      • Complete xx square: add (12×2)2=1(\frac{1}{2} \times 2)^2 = 1

      • Complete yy square: add [12×(−6)]2=9[\frac{1}{2} \times (-6)]^2 = 9

      • (x2+2x+1)+(y2−6y+9)=−7+1+9(x^2 + 2x + 1) + (y^2 - 6y + 9) = -7 + 1 + 9

      • (x+1)2+(y−3)2=3(x + 1)^2 + (y - 3)^2 = 3

    • This equation represents a circle with center (−1,3)(-1, 3) and radius r=3r = \sqrt{3}.

Plane Curves and Parametric Equations

  • Definitions

    • If ff and gg are functions defined on an interval II, the set of points (f(t),g(t))(f(t), g(t)) comprises a plane curve.

    • The equations expressing the coordinates as functions of a parameter tt are called parametric equations:     x=f(t)x = f(t)     y=g(t)y = g(t)

  • Example 14: Sketching a curve using parametric equations:

    • For x=t2−3tx = t^2 - 3t and y=t−1y = t - 1, a table of values can be generated:

      • If t=−2t = -2, x=10,y=−3x = 10, y = -3

      • If t=−1t = -1, x=4,y=−2x = 4, y = -2

      • If t=0t = 0, x=0,y=−1x = 0, y = -1

      • If t=1t = 1, x=−2,y=0x = -2, y = 0

      • If t=2t = 2, x=−2,y=1x = -2, y = 1

      • If t=3t = 3, x=0,y=2x = 0, y = 2

      • If t=4t = 4, x=4,y=3x = 4, y = 3

      • If t=5t = 5, x=10,y=4x = 10, y = 4

  • Example 15: Eliminating the parameter:

    • To convert the parametric equations from Example 14 into a rectangular (Cartesian) equation:

      1. Solve the simpler equation for tt: From y=t−1y = t - 1, we find t=y+1t = y + 1.

      2. Substitute into the second equation: x=(y+1)2−3(y+1)x = (y + 1)^2 - 3(y + 1).

      3. Expand and simplify:             x=(y2+2y+1)−(3y+3)x = (y^2 + 2y + 1) - (3y + 3)             x=y2−y−2x = y^2 - y - 2

    • The rectangular equation x=y2−y−2x = y^2 - y - 2 identifies the curve as a parabola.