Points, Lines & Circles

LessonAbout 1.5 h1 of 2 in Analytic Geometry 1

Analytic Geometry: Points, Lines, and Circles

Rectangular Coordinates, Distance, and Line Segments

Analytic geometry uses a rectangular coordinate system to convert geometric problems into algebraic equations. Points in a two-dimensional plane are defined by an abscissa (\(x\)-coordinate) and an ordinate (\(y\)-coordinate), while points in three-dimensional space include a \(z\)-coordinate.

Distance between two points in 2D space:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Distance between two points in 3D space:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
Coordinates of midpoint \(P_m(x_m, y_m)\):
\[ x_m = \dfrac{x_1 + x_2}{2}, \qquad\qquad y_m = \dfrac{y_1 + y_2}{2} \]
Division of line segment in ratio \(r_1 : r_2\):
\[ x = \dfrac{x_1 r_2 + x_2 r_1}{r_1 + r_2}, \qquad\qquad y = \dfrac{y_1 r_2 + y_2 r_1}{r_1 + r_2} \]
Area of a polygon by coordinates (Shoelace Formula):
\[ A = \dfrac{1}{2} |(x_1 y_2 + x_2 y_3 + \dots + x_n y_1) - (y_1 x_2 + y_2 x_3 + \dots + y_n x_1)| \]
Centroid of a triangle \((\bar{x}, \bar{y})\):
\[ \bar{x} = \dfrac{x_1 + x_2 + x_3}{3}, \qquad\qquad \bar{y} = \dfrac{y_1 + y_2 + y_3}{3} \]
  • Abscissa measures horizontal distance from y-axis
  • Ordinate measures vertical distance from x-axis
  • Midpoint bisects segment into two equal segments
  • Shoelace formula computes polygon area from ordered vertices

Worked Example 1: Distance between two points in 2D

Find the distance between the points \(A(4, -3)\) and \(B(-2, 5)\).

Solution: Apply the 2D distance formula:

\[ d = \sqrt{(-2 - 4)^2 + (5 - (-3))^2} = \sqrt{(-6)^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\text{ units} \]

Worked Example 2: Division of a line segment

Determine the coordinates of the point which is three-fifths of the way from \(P_1(2, -5)\) to \(P_2(-3, 5)\).

Solution: The ratio of segment lengths is \(r_1 = 3/5\) and \(r_2 = 2/5\). Apply the division formula:

\[ x = \dfrac{2(2/5) + (-3)(3/5)}{3/5 + 2/5} = \dfrac{4/5 - 9/5}{1} = -1 \]
\[ y = \dfrac{-5(2/5) + 5(3/5)}{3/5 + 2/5} = \dfrac{-10/5 + 15/5}{1} = 1 \]

The coordinates of the point are \((-1, 1)\).

Worked Example 3: Area of a triangle by coordinates

Find the area of a triangle with vertices at \((-2, 0)\), \((4, 0)\), and \((3, 3)\).

Solution: Apply the Shoelace formula using the given vertex coordinates:

\[ A = \dfrac{1}{2} |(-2)(0) + (4)(3) + (3)(0) - (0)(4) - (0)(3) - (3)(-2)| \]
\[ A = \dfrac{1}{2} |12 - (-6)| = \dfrac{18}{2} = 9\text{ sq. units} \]

Worked Example 4: Distance between two points in 3D space

Calculate the distance between points \(P_1(6, -2, -3)\) and \(P_2(5, 1, -4)\) in three-dimensional space.

Solution: Apply the 3D distance formula:

\[ d = \sqrt{(5 - 6)^2 + (1 - (-2))^2 + (-4 - (-3))^2} = \sqrt{(-1)^2 + 3^2 + (-1)^2} = \sqrt{1 + 9 + 1} = \sqrt{11} \]

Equations and Properties of Straight Lines

A straight line in a plane can be defined by its inclination angle, slope, or linear equations in various standard forms.

Slope of a line passing through \((x_1, y_1)\) and \((x_2, y_2)\):
\[ m = \tan\theta = \dfrac{y_2 - y_1}{x_2 - x_1} \]
General Equation of a line:
\[ Ax + By + C = 0 \]
Point-Slope Form:
\[ y - y_1 = m(x - x_1) \]
Slope-Intercept Form:
\[ y = mx + b \]
Two-Point Form:
\[ y - y_1 = \dfrac{y_2 - y_1}{x_2 - x_1}(x - x_1) \]
Intercept Form:
\[ \dfrac{x}{a} + \dfrac{y}{b} = 1 \]
Parallel Lines Condition:
\[ m_1 = m_2 \]
Perpendicular Lines Condition:
\[ m_2 = -\dfrac{1}{m_1} \]
Angle between two lines:
\[ \tan\theta = \dfrac{m_2 - m_1}{1 + m_1 m_2} \]
Distance from point \((x_1, y_1)\) to line \(Ax + By + C = 0\):
\[ d = \dfrac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
Distance between parallel lines \(Ax + By + C_1 = 0\) and \(Ax + By + C_2 = 0\):
\[ d = \dfrac{|C_1 - C_2|}{\sqrt{A^2 + B^2}} \]
  • Slope equals tangent of inclination angle with x-axis
  • Parallel lines have identical slope values
  • Perpendicular lines have negative reciprocal slopes
  • Distance from point to line is perpendicular length

Worked Example 5: Angle between two intersecting lines

Find the acute angle formed by the lines \(2x + y - 8 = 0\) and \(x + 3y + 4 = 0\).

Solution: Express both equations in slope-intercept form to find their slopes:

For line 1: \(y = -2x + 8 \implies m_1 = -2\).

For line 2: \(3y = -x - 4 \implies y = -\dfrac{1}{3}x - \dfrac{4}{3} \implies m_2 = -\dfrac{1}{3}\).

Apply the angle formula between two lines:

\[ \tan\theta = \dfrac{-1/3 - (-2)}{1 + (-2)(-1/3)} = \dfrac{5/3}{1 + 2/3} = \dfrac{5/3}{5/3} = 1 \]
\[ \theta = \arctan(1) = 45^\circ \]

Worked Example 6: Perpendicular distance from a point to a line

Determine the distance from the point \((1, 3)\) to the line \(4x + 3y + 12 = 0\).

Solution: Substitute \(x_1 = 1\), \(y_1 = 3\), \(A = 4\), \(B = 3\), and \(C = 12\) into the distance formula:

\[ d = \dfrac{|4(1) + 3(3) + 12|}{\sqrt{4^2 + 3^2}} = \dfrac{|4 + 9 + 12|}{\sqrt{25}} = \dfrac{25}{5} = 5\text{ units} \]

Worked Example 7: Distance between two parallel lines

Find the distance between the parallel lines \(4x - 3y = 12\) and \(4x - 3y = -8\).

Solution: Rewrite both equations in standard form \(Ax + By + C = 0\):

Line 1: \(4x - 3y - 12 = 0 \implies C_1 = -12\).

Line 2: \(4x - 3y + 8 = 0 \implies C_2 = 8\).

\[ d = \dfrac{|8 - (-12)|}{\sqrt{4^2 + (-3)^2}} = \dfrac{20}{5} = 4\text{ units} \]

Worked Example 8: Perpendicular lines coefficient determination

Determine the value of \(B\) such that the line \(3x + 2y - 7 = 0\) is perpendicular to \(2x - By + 2 = 0\).

Solution: The slope of the first line is \(m_1 = -3/2\). For the lines to be perpendicular, the slope \(m_2\) of the second line must be the negative reciprocal:

\[ m_2 = -\dfrac{1}{-3/2} = \dfrac{2}{3} \]

From the second line equation \(By = 2x + 2 \implies m_2 = \dfrac{2}{B}\). Equating slopes:

\[ \dfrac{2}{B} = \dfrac{2}{3} \implies B = 3 \]

Equations and Properties of Circles

A circle is the locus of all points in a plane that are equidistant from a fixed center point. Its equation can be written in standard or general form.

Standard Equation of a circle centered at \((h, k)\) with radius \(r\):
\[ (x - h)^2 + (y - k)^2 = r^2 \]
General Equation of a circle:
\[ Ax^2 + Ay^2 + Dx + Ey + F = 0 \]
Center \((h, k)\) and radius \(r\) formulas:
\[ h = -\dfrac{D}{2A}, \qquad\qquad k = -\dfrac{E}{2A}, \qquad\qquad r = \sqrt{\dfrac{D^2 + E^2 - 4AF}{4A^2}} \]
Point Circle Condition (\(r = 0\)):
\[ D^2 + E^2 - 4AF = 0 \]
Shortest distance from point \(P\) to circle with center \(C\) and radius \(r\):
\[ d_{\text{shortest}} = d_{PC} - r \]
Polar equation of circle centered at \((a, 0)\) with radius \(a\):
\[ r = 2a\cos\theta \]
  • Circle is locus of points equidistant from center
  • Standard form clearly gives center and radius
  • Completing the square converts general to standard form
  • Point circle occurs when radius equals zero

Worked Example 9: Finding center and radius by completing the square

Find the center coordinates and radius of the circle described by \(x^2 + y^2 - 6x + 4y - 12 = 0\).

Solution: Group \(x\) and \(y\) terms and complete the square for each variable:

\[ (x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4 \]
\[ (x - 3)^2 + (y + 2)^2 = 25 = 5^2 \]

The center is at \((3, -2)\) and the radius is \(r = 5\).

Worked Example 10: Shortest distance from a point to a circle

Calculate the shortest distance from the point \(A(3, 8)\) to the circle \(x^2 + y^2 + 4x - 6y = 12\).

Solution: Convert the circle equation to standard form to find its center and radius:

\[ (x + 2)^2 + (y - 3)^2 = 12 + 4 + 9 = 25 = 5^2 \]

The center is \(C(-2, 3)\) and radius \(r = 5\). Calculate distance from \(A(3, 8)\) to center \(C(-2, 3)\):

\[ d_{PC} = \sqrt{(3 - (-2))^2 + (8 - 3)^2} = \sqrt{5^2 + 5^2} = \sqrt{50} \approx 7.07\text{ units} \]

Subtract radius to find shortest distance to circumference:

\[ d_{\text{shortest}} = 7.07 - 5 = 2.07\text{ units} \]

Worked Example 11: Point circle parameter evaluation

Find the value of \(k\) for which the equation \(x^2 + y^2 + 4x - 2y - k = 0\) represents a point circle.

Solution: Complete the square to determine \(r^2\) as a function of \(k\):

\[ (x + 2)^2 + (y - 1)^2 = k + 2^2 + 1^2 = k + 5 \]

A point circle has radius \(r = 0\), so \(r^2 = 0\):

\[ k + 5 = 0 \implies k = -5 \]

Worked Example 12: Equation of a circle tangent to a line

Find the equation of the circle centered at \((-4, -5)\) and tangent to the line \(2x + 7y - 10 = 0\).

Solution: The radius \(r\) equals the perpendicular distance from center \((-4, -5)\) to the tangent line:

\[ r = \dfrac{|2(-4) + 7(-5) - 10|}{\sqrt{2^2 + 7^2}} = \dfrac{|-8 - 35 - 10|}{\sqrt{53}} = \dfrac{53}{\sqrt{53}} = \sqrt{53} \]

Substitute center \((-4, -5)\) and radius \(r = \sqrt{53}\) into the standard circle equation:

\[ (x + 4)^2 + (y + 5)^2 = (\sqrt{53})^2 \implies x^2 + 8x + 16 + y^2 + 10y + 25 = 53 \]
\[ x^2 + y^2 + 8x + 10y - 12 = 0 \]

Formula Summary: Analytic Geometry (Points, Lines, and Circles)

Formula Name Formula
2D Distance Formula \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
3D Distance Formula \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\)
Midpoint Formula \(x_m = \dfrac{x_1 + x_2}{2}, \quad y_m = \dfrac{y_1 + y_2}{2}\)
Line Segment Division \(x = \dfrac{x_1 r_2 + x_2 r_1}{r_1 + r_2}, \quad y = \dfrac{y_1 r_2 + y_2 r_1}{r_1 + r_2}\)
Shoelace Polygon Area \(A = \dfrac{1}{2} |(x_1 y_2 + x_2 y_3 + \dots + x_n y_1) - (y_1 x_2 + y_2 x_3 + \dots + y_n x_1)|\)
Triangle Centroid \(\bar{x} = \dfrac{x_1 + x_2 + x_3}{3}, \quad \bar{y} = \dfrac{y_1 + y_2 + y_3}{3}\)
Line Slope \(m = \tan\theta = \dfrac{y_2 - y_1}{x_2 - x_1}\)
General Line Equation \(Ax + By + C = 0\)
Point-Slope Form \(y - y_1 = m(x - x_1)\)
Slope-Intercept Form \(y = mx + b\)
Two-Point Form \(y - y_1 = \dfrac{y_2 - y_1}{x_2 - x_1}(x - x_1)\)
Intercept Form \(\dfrac{x}{a} + \dfrac{y}{b} = 1\)
Parallel Lines Condition \(m_1 = m_2\)
Perpendicular Lines Condition \(m_2 = -\dfrac{1}{m_1}\)
Angle Between Two Lines \(\tan\theta = \dfrac{m_2 - m_1}{1 + m_1 m_2}\)
Distance from Point to Line \(d = \dfrac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\)
Distance Between Parallel Lines \(d = \dfrac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}\)
Standard Circle Equation \((x - h)^2 + (y - k)^2 = r^2\)
General Circle Equation \(Ax^2 + Ay^2 + Dx + Ey + F = 0\)
Circle Center and Radius \(h = -\dfrac{D}{2A}, \quad k = -\dfrac{E}{2A}, \quad r = \sqrt{\dfrac{D^2 + E^2 - 4AF}{4A^2}}\)
Shortest Distance to Circle \(d_{\text{shortest}} = d_{PC} - r\)