Cartesian System of Rectangular Coordinates

Cartesian System of Rectangular Coordinates

• Definition and Scope of Coordinate Geometry:

  • Coordinate Geometry (also called Cartesian Geometry) is the branch of Mathematics that investigates the interrelationship between geometrical concepts and algebraic concepts.
  • Named in honor of the French mathematician René Descartes.
  • Geometrical locations are represented algebraically through numbers and letters (e.g., seat $H\text{-}4$ in a hall representing the fourth seat in row $H$).
  • Conversely, algebraic equations are represented geometrically via figures such as straight lines, circles, and polygons on maps.

• Rectangular Coordinate Axes:

  • The position of any point in a plane is fixed relative to two mutually perpendicular lines intersecting at a fixed reference point.
  • Origin ($O$): The point of intersection of the coordinate axes, assigned coordinates $O(0, 0)$.
  • x-axis ($\text{axis of } x$): The horizontal line $XOX'$.
  • y-axis ($\text{axis of } y$): The vertical line $YOY'$.

• Cartesian Coordinates of a Point:

  • To determine coordinates of a point $P$ in a plane, construct perpendiculars $PA \bot XOX'$ and $PB \bot YOY'$.
  • Abscissa ($x$-coordinate): The directed distance $OA = x$ measured along the x-axis.
  • Ordinate ($y$-coordinate): The directed distance $OB = PA = y$ measured along the y-axis.
  • Coordinates of $P$: Written as an ordered pair $(x, y)$.
  • Ordered Pair Property: Order is essential ($(x, y) \neq (y, x)$ unless $x = y$). For instance, $(3, 2)$ and $(2, 3)$ represent distinct points.

• Quadrants and Sign Conventions:

  • The axes $XOX'$ and $YOY'$ divide the coordinate plane into four regions called quadrants:
  • I Quadrant: $x > 0$, $y > 0$
  • II Quadrant: $x < 0$, $y > 0$
  • III Quadrant: $x < 0$, $y < 0$
  • IV Quadrant: $x > 0$, $y < 0$

Distance Between Two Points

• Derivation of the Distance Formula:

  • Let $P(x_1, y_1)$ and $Q(x_2, y_2)$ be two points in the plane.
  • Draw horizontal line $l$ through $P$ parallel to $XX'$, and a vertical perpendicular from $Q$ meeting $l$ at $R$.
  • In right-angled triangle $\Delta PQR$:
  • $PR = x_2 - x_1$
  • $QR = y_2 - y_1$
  • Applying Pythagoras theorem:     $PQ^2 = PR^2 + QR^2$$PQ^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$
  • Distance Formula:     $PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
  • Holds universally for points in any quadrant.

• Distance from Origin:

  • Distance of $P(x, y)$ from origin $O(0, 0)$:     $OP = \sqrt{x^2 + y^2}$

• Worked Examples:

  • Pairwise Distance Calculations:
  • Points $A(14, 3)$ and $B(10, 6)$: $AB = \sqrt{(10 - 14)^2 + (6 - 3)^2} = \sqrt{(-4)^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5\,\text{units}$.
  • Points $M(-1, 2)$ and $N(0, -6)$: $MN = \sqrt{(0 - (-1))^2 + (-6 - 2)^2} = \sqrt{1^2 + (-8)^2} = \sqrt{1 + 64} = \sqrt{65}\,\text{units}$.
  • Verification of Right-Angled Triangle:
  • Vertices $P(-1, -1)$, $Q(2, 3)$, $R(-2, 6)$:       $PQ^2 = (2 + 1)^2 + (3 + 1)^2 = 3^2 + 4^2 = 25$$QR^2 = (-2 - 2)^2 + (6 - 3)^2 = (-4)^2 + 3^2 = 25$$RP^2 = (-1 - (-2))^2 + (-1 - 6)^2 = 1^2 + (-7)^2 = 50$$PQ^2 + QR^2 = 25 + 25 = 50 = RP^2 \implies \Delta PQR$ is right-angled by converse of Pythagoras Theorem.
  • Collinearity Verification:
  • Points $A(1, 2)$, $B(4, 5)$, $C(-1, 0)$:       $AB = \sqrt{(4 - 1)^2 + (5 - 2)^2} = \sqrt{18} = 3\sqrt{2}\,\text{units}$$BC = \sqrt{(-1 - 4)^2 + (0 - 5)^2} = \sqrt{50} = 5\sqrt{2}\,\text{units}$$AC = \sqrt{(-1 - 1)^2 + (0 - 2)^2} = \sqrt{8} = 2\sqrt{2}\,\text{units}$$AB + AC = 3\sqrt{2} + 2\sqrt{2} = 5\sqrt{2} = BC \implies A, B, C$ are collinear.
  • Equilateral Triangle Verification:
  • Vertices $A(2a, 4a)$, $B(2a, 6a)$, $C(2a + \sqrt{3}a, 5a)$.       $AB = \sqrt{(2a - 2a)^2 + (6a - 4a)^2} = 2a\,\text{units}$$BC = \sqrt{(2a + \sqrt{3}a - 2a)^2 + (5a - 6a)^2} = \sqrt{3a^2 + a^2} = 2a\,\text{units}$$AC = \sqrt{(2a + \sqrt{3}a - 2a)^2 + (5a - 4a)^2} = \sqrt{3a^2 + a^2} = 2a\,\text{units}$$AB = BC = AC = 2a \implies$ Vertices form an equilateral triangle of side length $2a$.

Section Formula

• Internal Division:

  • Let $R(x, y)$ divide the line segment joining $P(x_1, y_1)$ and $Q(x_2, y_2)$ internally in ratio $m_1 : m_2$.
  • By similar triangles ($\Delta RPS \sim \Delta QRT$):     $\frac{m_1}{m_2} = \frac{x - x_1}{x_2 - x} = \frac{y - y_1}{y_2 - y}$
  • Coordinates of $R$:     $x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}$$y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}$
  • Expressed as ordered pair:     $R\left(\frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}, \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}\right)$

• Coordinates of Mid-Point:

  • If $R$ is the mid-point of $PQ$, $m_1 : m_2 = 1 : 1$:     $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$

• External Division:

  • Let $R(x, y)$ divide the line segment joining $P(x_1, y_1)$ and $Q(x_2, y_2)$ externally in ratio $m_1 : m_2$.
  • Coordinates of $R$:     $x = \frac{m_1 x_2 - m_2 x_1}{m_1 - m_2}$$y = \frac{m_1 y_2 - m_2 y_1}{m_1 - m_2}$
  • Expressed as ordered pair:     $R\left(\frac{m_1 x_2 - m_2 x_1}{m_1 - m_2}, \frac{m_1 y_2 - m_2 y_1}{m_1 - m_2}\right)$

• Worked Examples:

  • Internal and External Points Calculation:
  • Segment joining $(4, -2)$ and $(-3, 5)$ in ratio $2 : 3$:
    • Internal Point $P$: $x = \frac{2(-3) + 3(4)}{2 + 3} = \frac{6}{5}$, $y = \frac{2(5) + 3(-2)}{2 + 3} = \frac{4}{5} \implies P\left(\frac{6}{5}, \frac{4}{5}\right)$.
    • External Point $Q$: $x' = \frac{2(-3) - 3(4)}{2 - 3} = 18$, $y' = \frac{2(5) - 3(-2)}{2 - 3} = -16 \implies Q(18, -16)$.
  • Determining Ratio of Division:
  • Point $P(3, -2)$ dividing segment joining $(1, 4)$ and $(-3, 16)$ in ratio $k : 1$:       $\frac{-3k + 1}{k + 1} = 3 \implies -3k + 1 = 3k + 3 \implies 6k = -2 \implies k = -\frac{1}{3}$       Negative value signifies external division in the ratio 1 : 3$.\n - **Parallelogram Midpoint Proof**:\n - Vertices: A(1, 4)$,$B(-2, 1)$,$C(0, -1)$,$D(3, 2).\n - Midpoints E, F, G, H$of sides$AB, BC, CD, DA:\n      E = \left(-\frac{1}{2}, \frac{5}{2}\right)$,$F = \left(-1, 0\right)$,$G = \left(\frac{3}{2}, \frac{1}{2}\right)$,$H = \left(1, 3\right).\n - Midpoint of diagonal EG$:$\left(\frac{-1/2 + 3/2}{2}, \frac{5/2 + 1/2}{2}\right) = \left(\frac{1}{2}, \frac{3}{2}\right).\n - Midpoint of diagonal FH$:$\left(\frac{-1 + 2}{2}, \frac{0 + 3}{2}\right) = \left(\frac{1}{2}, \frac{3}{2}\right).\n - Diagonals share the same midpoint, so they bisect each other, proving EFGH is a parallelogram.\n\n# Area of a Triangle\n\n- **Geometric Derivation**:\n - For vertices A(x_1, y_1)$,$B(x_2, y_2)$,$C(x_3, y_3)$, draw perpendiculars$AL, BM, CN to x-axis.\n - \text{Area of } \Delta ABC = \text{Area}(BMLA) + \text{Area}(ALNC) - \text{Area}(BMNC)\n - Expansion formula:\n    \text{Area of } \Delta ABC = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|\n\n- **Determinant Form**:\n  \text{Area of } \Delta ABC = \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \ x_2 & y_2 & 1 \ x_3 & y_3 & 1 \end{vmatrix}\n\n- **Worked Examples**:\n - **Area Calculation**:\n - Vertices A(3, 4)$,$B(6, -2)$,$C(-4, -5).\n      \text{Area} = \frac{1}{2} |3(-2 - (-5)) + 6(-5 - 4) + (-4)(4 - (-2))|\n      \text{Area} = \frac{1}{2} |3(3) + 6(-9) + (-4)(6)| = \frac{1}{2} |9 - 54 - 24| = \frac{1}{2} |-69| = \frac{69}{2}\,\text{sq. units}\n - **Solving Unknown Parameter via Given Area**:\n - Vertices (1, k)$,$(4, -3)$,$(-9, 7)$with Area =$15\,\text{sq. units}.\n      \frac{1}{2} |1(-3 - 7) + 4(7 - k) + (-9)(k - (-3))| = 15\n      \frac{1}{2} |-10 + 28 - 4k - 9k - 27| = 15 \implies \frac{1}{2} |-9 - 13k| = 15\n      -9 - 13k = 30 \implies -13k = 39 \implies k = -3\n\n# Condition for Collinearity of Three Points\n\n- **Collinearity Condition**:\n - Three points A(x_1, y_1)$,$B(x_2, y_2)$,$C(x_3, y_3)$are collinear if and only if the area of the triangle formed by them is zero ($\text{Area} = 0).\n - Algebraic Form:\n    x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 0\n - Determinant Form:\n    \begin{vmatrix} x_1 & y_1 & 1 \ x_2 & y_2 & 1 \ x_3 & y_3 & 1 \end{vmatrix} = 0\n\n- **Worked Examples**:\n - **Collinearity Proof**:\n - Points A(a, b+c)$,$B(b, c+a)$,$C(c, a+b).\n      \begin{vmatrix} a & b+c & 1 \ b & c+a & 1 \ c & a+b & 1 \end{vmatrix}\n      Apply column transformation C_1 \rightarrow C_1 + C_2:\n      \begin{vmatrix} a+b+c & b+c & 1 \ a+b+c & c+a & 1 \ a+b+c & a+b & 1 \end{vmatrix} = (a+b+c) \begin{vmatrix} 1 & b+c & 1 \ 1 & c+a & 1 \ 1 & a+b & 1 \end{vmatrix} = (a+b+c)(0) = 0\n      Hence, points are collinear.\n - **Solving Parameter for Collinearity**:\n - Points (1, 5)$,$(k, 1)$,$(4, 11) are collinear.\n      \frac{1}{2} [1(1 - 11) + k(11 - 5) + 4(5 - 1)] = 0\n      \frac{1}{2} [-10 + 6k + 16] = 0 \implies 6k + 6 = 0 \implies k = -1\n\n# Inclination and Slope of a Line\n\n- **Inclination of a Line (\theta)**:\n - The measure of angle made by the line with the positive direction of the x-axis, measured in the anti-clockwise direction.\n - For lines parallel to or coinciding with the x-axis, inclination is defined as \theta = 0^\circ\n\n- **Slope of a Line (m)**:\n - Tangent of the inclination angle \theta:\n    m = \tan(\theta)\n - Undefined Slope: If a line makes 90^\circ$or$270^\circ with x-axis, slope is undefined.\n - Direction Independence: Slope of line segment AB$equals slope of$BA$($\tan(\pi + \alpha) = \tan(\alpha)).\n\n- **Special Cases**:\n - Line making 30^\circ$with negative direction of x-axis:$\theta = 180^\circ - 30^\circ = 150^\circ \implies m = \tan(150^\circ) = -\frac{1}{\sqrt{3}}.\n - Line making 60^\circ$with positive direction of y-axis:$\theta = 90^\circ + 60^\circ = 150^\circ \implies m = \tan(150^\circ) = -\cot(60^\circ) = -\frac{1}{\sqrt{3}}.\n - Lines equally inclined to coordinate axes: Inclinations are 45^\circ$or$135^\circ \implies m = \tan(45^\circ) = 1$or$m = \tan(135^\circ) = -1 \implies m = \pm 1$.

Slope of a Line Joining Two Distinct Points

• Two-Point Formula:

  • For points $A(x_1, y_1)$ and $B(x_2, y_2)$:     $m = \frac{y_2 - y_1}{x_2 - x_1}$
  • If $x_1 = x_2$, slope $m$ is undefined (line is parallel to y-axis).

• Worked Examples:

  • Slope between $A(6, 3)$ and $B(4, 10)$: $m = \frac{10 - 3}{4 - 6} = -\frac{7}{2}$.
  • Find $x$ if slope through $(3, 6)$ and $(x, 4)$ is $2$:     $\frac{4 - 6}{x - 3} = 2 \implies \frac{-2}{x - 3} = 2 \implies 2x - 6 = -2 \implies x = 2$

Conditions for Parallelism and Perpendicularity of Lines

• Parallel Lines:

  • Two non-vertical lines $l_1, l_2$ with slopes $m_1, m_2$ are parallel if and only if their slopes are equal:     $m_1 = m_2 \iff l_1 \parallel l_2$

• Perpendicular Lines:

  • Two non-vertical lines $l_1, l_2$ with slopes $m_1, m_2$ are perpendicular if and only if the product of their slopes is $-1$:     $m_1 \cdot m_2 = -1 \iff l_1 \bot l_2$

• Worked Examples:

  • Parallelism Verification:
  • Line through $A(5, 6)$ and $B(2, 3)$ has slope m_1 = \frac{3 - 6}{2 - 5} = 1$.\n - Line through C(9, -2)$and$D(6, -5)$has slope$m_2 = \frac{-5 - (-2)}{6 - 9} = 1$.
  • m_1 = m_2 = 1 \implies AB \parallel CD$.\n - **Perpendicularity Verification**:\n - Line through A(2, -5)$and$B(-2, 5)$has slope$m_1 = \frac{5 - (-5)}{-2 - 2} = -\frac{5}{2}$.
  • Line through $L(6, 3)$ and $M(1, 1)$ has slope m_2 = \frac{1 - 3}{1 - 6} = \frac{2}{5}$.\n - m_1 \cdot m_2 = \left(-\frac{5}{2}\right)\left(\frac{2}{5}\right) = -1 \implies AB \bot LM$.
  • Right-Angled Triangle via Slopes:
  • Vertices $A(4, 4)$, $B(3, 5)$, $C(-1, -1)$.       $m_{AB} = \frac{5 - 4}{3 - 4} = -1$$m_{BC} = \frac{-1 - 5}{-1 - 3} = \frac{3}{2}$$m_{AC} = \frac{-1 - 4}{-1 - 4} = 1$$m_{AB} \cdot m_{AC} = (-1)(1) = -1 \implies AC \bot AB \implies \Delta ABC$ is right-angled at A$.\n - **Solving Parameter via Perpendicularity**:\n - Line through A(3, y)$and$B(2, 7)$perpendicular to line through$C(-1, 4)$and$D(0, 6).\n      m_1 = \frac{7 - y}{2 - 3} = y - 7\n      m_2 = \frac{6 - 4}{0 - (-1)} = 2\n      m_1 \cdot m_2 = -1 \implies (y - 7)(2) = -1 \implies 2y - 14 = -1 \implies y = \frac{13}{2}\n\n# Intercepts Made by a Line on Axes\n\n- **Definitions**:\n - For line l$meeting x-axis at$A(a, 0)$and y-axis at$B(0, b):\n - **x-intercept (a$)**: Distance$OA$. Calculated by setting$y = 0 in line equation.\n - **y-intercept (b$or$c$)**: Distance$OB$. Calculated by setting$x = 0 in line equation.\n - **Intercepted Portion**: Segment AB between coordinate axes.\n - **Special Properties**:\n - A line passing through origin makes zero intercepts (a = 0, b = 0).\n - A horizontal line has no x-intercept.\n - A vertical line has no y-intercept.\n\n- **Worked Example**:\n - Line 2x + 3y = 6:\n - Set x = 0 \implies 3y = 6 \implies y = 2$(y-intercept is$2).\n - Set y = 0 \implies 2x = 6 \implies x = 3$(x-intercept is$3).\n\n# Angle Between Two Lines\n\n- **Formula Derivation**:\n - Let non-vertical lines l_1, l_2$have slopes$m_1 = \tan(\alpha_1)$and$m_2 = \tan(\alpha_2).\n - Exterior angle relation: \alpha_1 = \alpha_2 + \theta \implies \theta = \alpha_1 - \alpha_2$.
  • Tangent of angle between lines:     $\tan(\theta) = \tan(\alpha_1 - \alpha_2) = \frac{\tan(\alpha_1) - \tan(\alpha_2)}{1 + \tan(\alpha_1)\tan(\alpha_2)} = \frac{m_1 - m_2}{1 + m_1 m_2}$
  • Acute Angle ($\theta$):     $\tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|, \quad \text{where } 1 + m_1 m_2 \neq 0$
  • Obtuse Angle ($\phi$):     $\phi = 180^\circ - \theta, \quad \tan(\phi) = -\frac{m_1 - m_2}{1 + m_1 m_2}$

• Worked Examples:

  • Angle Calculation:
  • Slopes $m_1 = \frac{3}{4}$ and $m_2 = -\frac{1}{7}$:       $\tan(\theta) = \left| \frac{\frac{3}{4} - (-\frac{1}{7})}{1 + (\frac{3}{4})(-\frac{1}{7})} \right| = \left| \frac{\frac{25}{28}}{\frac{25}{28}} \right| = 1 \implies \theta = 45^\circ$       Obtuse angle $\phi = 180^\circ - 45^\circ = 135^\circ$
  • Angle Between x-axis and Line:
  • Line through $(3, -1)$ and $(4, -2)$: $m_1 = 0$ (x-axis), m_2 = \frac{-2 - (-1)}{4 - 3} = -1$.\n      \tan(\theta) = \left| \frac{0 - (-1)}{1 + (0)(-1)} \right| = 1 \implies \theta = 45^\circ\n - **Calculating Slope from Inter-line Angle**:\n - Angle = \frac{\pi}{4}$($45^\circ$),$m_1 = \frac{1}{2}.\n      \tan\left(\frac{\pi}{4}\right) = \left| \frac{m_2 - \frac{1}{2}}{1 + \frac{1}{2}m_2} \right| = 1 \implies \frac{2m_2 - 1}{2 + m_2} = \pm 1\n - Case 1: \frac{2m_2 - 1}{2 + m_2} = 1 \implies 2m_2 - 1 = 2 + m_2 \implies m_2 = 3\n - Case 2: \frac{2m_2 - 1}{2 + m_2} = -1 \implies 2m_2 - 1 = -2 - m_2 \implies 3m_2 = -1 \implies m_2 = -\frac{1}{3}\n      Slope of second line is 3$or$-\frac{1}{3}.\n\n# Shifting of Origin (Translation of Axes)\n\n- **Translation of Axes**:\n - Shifting origin to a new reference point O'(h, k) without altering directions of coordinate axes.\n\n- **Transformation Formulas**:\n - Point P(x, y)$in original system transforms to$P(x', y') in new system:\n    x = x' + h \implies x' = x - h\n    y = y' + k \implies y' = y - k\n - Equation transformation: Original curve equation F(x, y) = 0$transforms to$F(x' + h, y' + k) = 0\n - Holds universally across all quadrants.\n\n- **Worked Examples**:\n - **New Point Coordinates**:\n - Origin shifted to (-3, 2)$, original point$(1, 2):\n      x' = 1 - (-3) = 4\n      y' = 2 - 2 = 0\n      Transformed coordinates: (4, 0)\n - **Transformed Curve Equation**:\n - Origin shifted to (3, 4)$, line$3x + 2y - 5 = 0:\n      Substitute x = x' + 3$and$y = y' + 4:\n      3(x' + 3) + 2(y' + 4) - 5 = 0\n      3x' + 9 + 2y' + 8 - 5 = 0 \implies 3x' + 2y' + 12 = 0\n\n# Comprehensive Problem Set Solutions & Reference Answers\n\n- **Check Your Progress 13.1**:\n - 1(a)$Distance between$(5, 4)$and$(2, -3)$:$\sqrt{58}\,\text{units}.\n - 1(b)$Distance between$(a, -a)$and$(b, b)$:$\sqrt{2(a + b)^2} = \sqrt{2}(a + b)\,\text{units}.\n- **Check Your Progress 13.2**:\n - 1(a)$Midpoint of$(-2, 3)$and$(3, 5)$:$\left(\frac{1}{2}, 4\right).\n - 1(b)$Midpoint of$(6, 0)$and$(-2, 10)$:$(2, 5).\n - 2$Point dividing$(-5, -2)$and$(3, 6)$internally in ratio$3 : 1$:$(1, 4).\n - 3(a)$Fourth vertex of parallelogram with vertices$(0, 3)$,$(0, 6)$,$(2, 9)$:$(2, 6).\n - 4$Points of trisection of segment joining$(2, 3)$and$(5, -1)$:$\left(3, \frac{5}{3}\right)$and$\left(4, \frac{1}{3}\right).\n- **Check Your Progress 13.3**:\n - 1(a)$Area with vertices$(0, 5)$,$(5, -5)$,$(0, 0)$:$\frac{25}{2}\,\text{sq. units}.\n - 1(b)$Area with vertices$(2, 3)$,$(-2, -3)$,$(-2, 3)$:$12\,\text{sq. units}.\n - 1(c)$Area with vertices$(a, 0)$,$(0, -a)$,$(0, 0)$:$\frac{a^2}{2}\,\text{sq. units}.\n - 2$Value of$k$for vertices$A(2, -3)$,$B(3, -2)$,$C\left(\frac{5}{2}, k\right)$with area$\frac{3}{2}\,\text{sq. units}$:$k = 3$or$k = 5.\n - 3$Area of rectangle with vertices$(5, 4)$,$(5, -4)$,$(-5, 4)$,$(-5, -4)$:$80\,\text{sq. units}.\n - 4$Area of quadrilateral with vertices$(5, -2)$,$(4, -7)$,$(1, 1)$,$(3, 4)$:$\frac{41}{2}\,\text{sq. units}.\n- **Check Your Progress 13.4**:\n - 3$Condition for$(a, 0)$,$(0, b)$,$(1, 1)$collinearity:$\frac{1}{a} + \frac{1}{b} = 1$.
  • $5$ Value of $k$ for collinear points $(5, 7)$, $(k, 5)$, $(0, 2)$: k = 3$.\n - 6$Values of$k$for collinear points$(k, 2-2k)$,$(-k+1, 2k)$,$(-4-k, 6-2k)$:$k = \frac{1}{2}$or$k = -1$.

• Check Your Progress 13.5:

  • $1(i)$ Slope at inclination $60^\circ$: $\sqrt{3}$.
  • $1(ii)$ Slope at inclination $150^\circ$: $-\frac{1}{\sqrt{3}}$.
  • $2$ Slope at $30^\circ$ with positive y-axis: $-\sqrt{3}$.
  • $3$ Slope at $60^\circ$ with negative x-axis: $-\sqrt{3}$.

• Check Your Progress 13.6:

  • $1$ Slope of line through $A(6, 8)$ and $B(4, 14)$: -3$.\n - 2$Value of$x$for slope$4$through$A(6, 12)$and$B(x, 8)$:$x = 5$.
  • $3$ Value of $y$ for slope $\frac{4}{3}$ through $A(-8, 11)$ and $B(2, y)$: $y = -\frac{7}{3}$.
  • $4$ Slope of line passing through $B(0, 4)$ and midpoint of $AC$ where $A(2, 3)$, $C(-5, 0)$: \text{Midpoint} = \left(-\frac{3}{2}, \frac{3}{2}\right) \implies m = \frac{5}{3}$.\n- **Check Your Progress 13.7**:\n - 2$Slope of line parallel to line joining$(-4, 1)$and$(2, 3)$:$m = \frac{3 - 1}{2 - (-4)} = \frac{1}{3}$.
  • $3$ Value of $x$ if line through $(-5, 7)$ and $(0, -2)$ is perpendicular to line through $(1, 3)$ and $(4, x)$: m_1 = -\frac{9}{5} \implies m_2 = \frac{5}{9} \implies \frac{x - 3}{3} = \frac{5}{9} \implies x = \frac{14}{3}$.\n - 6$Value of$k$for parallel lines through$(k, 9)$,$(2, 7)$and$(2, -2)$,$(6, 4)$:$m_1 = \frac{-2}{2 - k}$,$m_2 = \frac{6}{4} = \frac{3}{2} \implies \frac{-2}{2 - k} = \frac{3}{2} \implies k = \frac{10}{3}$.

• Check Your Progress 13.8:

  • $1(i)$ Intercepts of $x + 3y = 6$: x-intercept = $6$, y-intercept = 2$.\n - 1(ii)$Intercepts of$2x + 3y = 7$: x-intercept =$\frac{7}{2}$, y-intercept =$\frac{7}{3}$.
  • $1(iii)$ Intercepts of $\frac{x}{2a} + \frac{y}{2b} = 1$: x-intercept = $2a$, y-intercept = 2b$.\n - 1(iv)$Intercepts of$ax + by = c$: x-intercept =$\frac{c}{a}$, y-intercept =$\frac{c}{b}$.
  • $1(v)$ Intercepts of $2x = 8 - 2y \implies x + y = 4$: x-intercept = $-4$, y-intercept = 16$.\n - 1(vi)$Intercepts of$2x - 3y = 7 \cdot 3 = 21$: x-intercept =$-\frac{21}{2}$, y-intercept =$21$.

• Check Your Progress 13.9:

  • $1$ Acute angle between slopes $5$ and $\frac{2}{3}$: $45^\circ$
  • $2$ Obtuse angle between slopes $2$ and $-3$: $135^\circ$
  • $3$ Acute angle between line $l_1$ through $(0, 0)$, $(2, 3)$ ($m_1 = \frac{3}{2}$) and $l_2$ through $(2, -2)$, $(3, 5)$ ($m_2 = 7$): \tan(\theta) = \left| \frac{7 - 3/2}{1 + 21/2} \right| = \frac{11}{23} \implies \theta = \tan^{-1}\left(\frac{11}{23}\right)$.\n- **Check Your Progress 13.10**:\n - 1(i) Length change due to axis translation: No.\n - 1(ii) Fixed points under translation: No.\n - 1(iii)$Point$(0, 3)$coordinates when origin shifted to$(4, -5)$:$(-4, 8).\n - 1(iv)$Original coordinates of$P$if shifted point is$(4, 5)$under shift to$(2, 3)$:$(6, 8).\n - 1(v)$New origin location if$(3, 0)$changes to$(2, -3)$:$(1, 3).\n- **Terminal Exercise Answers**:\n - 1(a)$Distance between$(2, 0)$and$(1, \cot(\theta))$:$\csc(\theta).\n - 1(b)$Distance between$(-\sin(A), \cos(A))$and$(\sin(B), \cos(B))$:$2\sin\left(\frac{A + B}{2}\right).\n - 2$Triangle formation check: Neither$(3, 2), (-3, 2), (0, 3)$nor$(3, 2), (3, -2), (3, 0) form a triangle.\n - 3$Midpoint of$(3, -5)$and$(-6, 8)$:$\left(-\frac{3}{2}, \frac{3}{2}\right).\n - 4(a)$Area of triangle with vertices$(1, 2)$,$(-2, 3)$,$(-3, -4)$:$11\,\text{sq. units}.\n - 4(b)$Area of triangle with vertices$(c, a)$,$(c + a, a)$,$(c - a, -a)$:$a^2\,\text{sq. units}.\n - 6$Value of$x$if$(-3, 12)$,$(7, 6)$,$(x, a)$are collinear:$x = \frac{51 - 5a}{3}.\n - 7$Area of quadrilateral with vertices$(4, 3)$,$(-5, 6)$,$(0, 7)$,$(3, -6)$:$29\,\text{sq. units}.\n - 8(a)$Slope through$(1, 2)$and$(4, 2)$:$0$.
  • $8(b)$ Slope through $(4, -6)$ and $(-2, -5)$: -\frac{1}{6}$.\n - 9$Value of$y$so line through$(3, y)$and$(2, 7)$is parallel to line through$(-1, 4)$and$(0, 6)$:$y = 9$.
  • $11$ Collinearity check using slopes: Only set $(ii)$ $(5, 1), (1, -1), (11, 4)$ is collinear.
  • $12$ Slopes for rectangle vertices $A(2, -3)$ and $B(3, 5)$: $\text{Slope of } AB = 8 \implies$ (i) $\text{Slope of } BC = -\frac{1}{8}$, (ii) $\text{Slope of } CD = 8$, (iii) \text{Slope of } DA = -\frac{1}{8}$.\n - 14$x-intercepts: (i)$3x - 2y = 8 \implies 4$, (ii)$7x + 3y - 9 = 0 \implies -3$, (iii)$y = 2x - 3 \implies \frac{3}{2}$.
  • $15$ Transformed equation of $x^2 + 2xy + y^2 + 4x + 5y = 0$ under shift to $(3, 4)$: $x'^2 + 2x'y' + y'^2 + 18x' + 15y' + 83 = 0$.
  • $16$ Original equation if transformed equation is $\frac{1}{2}(x')^2 + \frac{1}{2}(y')^2 = 4$ under shift to $(3, -4)$: x^2 + y^2 - 6x + 8y + 17 = 0$.\n - 17$Angle$\angle ABC$for$A(-2, 3)$,$B(3, 8)$,$C(4, 1)$:$\tan^{-1}\left(\frac{4}{3}\right)$.
  • $18$ Acute angle between diagonals of quadrilateral $ABCD$ with $A(9, 2)$, $B(17, 11)$, $C(5, -3)$, $D(-3, -2)$: \tan^{-1}\left(\frac{48}{145}\right)$.\n - 19$Acute angle between lines$AB$and$BC$for$A(5, -3)$,$B(-3, -2)$,$C(9, 12)$:$\tan^{-1}\left(\frac{62}{55}\right)$.