Coordinate Geometry: Linear Equations, Slopes, and Intercepts

Fundamental Concepts and Formulas of Linear Equations

  • Definition of Slope: The slope mm measures the steepness and direction of a non-vertical line in the Cartesian coordinate plane. It is defined as the ratio of the vertical change (rise, Δy\Delta y) to the horizontal change (run, Δx\Delta x):

    • m=ΔyΔx=y2−y1x2−x1m = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1}
    • A positive slope (m>0m > 0) indicates that the line rises from left to right.
    • A negative slope (m<0m < 0) indicates that the line falls from left to right.
    • A zero slope (m=0m = 0) represents a horizontal line (y=cy = c).
    • An undefined slope represents a vertical line (x=cx = c).

  • Forms of Linear Equations:

    • Slope-Intercept Form: Expressed as y=mx+by = mx + b, where mm denotes the slope and bb denotes the yy-intercept (the point (0,b)(0, b)). This form allows immediate identification of the line's rate of change and vertical starting value.
    • Point-Slope Form: Expressed as y−y1=m(x−x1)y - y_1 = m(x - x_1), where mm is the slope and (x1,y1)(x_1, y_1) is any known coordinate point on the line. This form is particularly useful when given a point and a slope, or two points.
    • Standard Form: Expressed as Ax+By=CAx + By = C, where AA, BB, and CC are integers (typically with A≥0A \ge 0).

  • Intercepts of a Line:

    • xx-Intercept: The coordinate point where the graph crosses the xx-axis. At this point, the vertical coordinate is zero (y=0y = 0). An xx-intercept of aa corresponds to the ordered pair (a,0)(a, 0).
    • yy-Intercept: The coordinate point where the graph crosses the yy-axis. At this point, the horizontal coordinate is zero (x=0x = 0). A yy-intercept of bb corresponds to the ordered pair (0,b)(0, b).

  • Parallel and Perpendicular Line Relationships:

    • Parallel Lines: Two non-vertical lines in the same plane are parallel if and only if they possess identical slopes and distinct yy-intercepts:
    • m1=m2m_1 = m_2
    • b1≠b2b_1 \neq b_2

  • Perpendicular Lines: Two non-vertical lines are perpendicular if and only if their slopes are negative reciprocals of one another (meaning their product equals −1-1):

    • m1×m2=−1m_1 \times m_2 = -1
    • m2=−1m1m_2 = -\frac{1}{m_1}

Problem 6: Line Properties and True/False Statement Analysis

  • Problem Statement:

    • The line l1l_1 passes through the point (−4,−4)(-4, -4) and has an xx-intercept of −7-7. Which of the following are false?
    • I. l1l_1 has a yy-intercept of −7-7
    • II. l1l_1 passes through the point (8,2)(8, 2)
    • III. l1l_1 is perpendicular to y=34x−2y = \frac{3}{4}x - 2

  • Available Multiple-Choice Options:

    • (a) I and II only
    • (b) I and III only
    • (c) II and III only
    • (d) I, II, and III

  • Step-by-Step Derivation of Line l1l_1:

    • Identify the two given coordinate points on line l1l_1:
    • First point: P1=(−4,−4)P_1 = (-4, -4)
    • Second point derived from the xx-intercept of −7-7: P2=(−7,0)P_2 = (-7, 0)
    • Calculate the slope m1m_1 using the slope formula:
    • m1=y2−y1x2−x1m_1 = \frac{y_2 - y_1}{x_2 - x_1}
    • m1=0−(−4)−7−(−4)m_1 = \frac{0 - (-4)}{-7 - (-4)}
    • m1=4−7+4m_1 = \frac{4}{-7 + 4}
    • m1=4−3=−43m_1 = \frac{4}{-3} = -\frac{4}{3}
    • Determine the equation of line l1l_1 using point-slope form with (−7,0)(-7, 0):
    • y−0=−43(x−(−7))y - 0 = -\frac{4}{3}(x - (-7))
    • y=−43(x+7)y = -\frac{4}{3}(x + 7)
    • y=−43x−283y = -\frac{4}{3}x - \frac{28}{3}

  • Systematic Evaluation of Statements:

    • Evaluation of Statement I (l1l_1 has a yy-intercept of −7-7):
    • Determine the true yy-intercept by evaluating the equation at x=0x = 0:
    • y=−43(0)−283=−283y = -\frac{4}{3}(0) - \frac{28}{3} = -\frac{28}{3}
    • In mixed number form, −283=−913≈−9.33-\frac{28}{3} = -9\frac{1}{3} \approx -9.33
    • Because −913≠−7-9\frac{1}{3} \neq -7, Statement I is FALSE.

  • Evaluation of Statement II (l1l_1 passes through the point (8,2)(8, 2)) :

    • Substitute x=8x = 8 into the equation of l1l_1:
    • y=−43(8)−283y = -\frac{4}{3}(8) - \frac{28}{3}
    • y=−323−283y = -\frac{32}{3} - \frac{28}{3}
    • y=−603=−20y = -\frac{60}{3} = -20
    • The actual coordinate corresponding to x=8x = 8 is (8,−20)(8, -20). Because −20≠2-20 \neq 2, the point (8,2)(8, 2) does not lie on line l1l_1. Thus, Statement II is FALSE.
    • Evaluation of Statement III (l1l_1 is perpendicular to y=34x−2y = \frac{3}{4}x - 2):
    • Identify the slope of the comparison line y=34x−2y = \frac{3}{4}x - 2:
    • m2=34m_2 = \frac{3}{4}
    • Check the perpendicularity condition m1×m2=−1m_1 \times m_2 = -1:
    • m1×m2=(−43)×(34)=−1212=−1m_1 \times m_2 = \left(-\frac{4}{3}\right) \times \left(\frac{3}{4}\right) = -\frac{12}{12} = -1
    • Because the product of the slopes is −1-1, line l1l_1 is indeed perpendicular to y=34x−2y = \frac{3}{4}x - 2. Thus, Statement III is TRUE.

  • Conclusion:

    • The statements that are false are Statement I and Statement II.
    • Correct Option: (a) I and II only

Problem 7: Constructing Parallel Linear Equations

  • Problem Statement:

    • What is the equation of the line that passes through the point (6,−1)(6, -1) and is parallel to the line y=23x+4y = \frac{2}{3}x + 4?

  • Available Multiple-Choice Options:

    • (a) y=23x−5y = \frac{2}{3}x - 5
    • (b) y=−32x−6y = -\frac{3}{2}x - 6
    • (c) y=23x+6y = \frac{2}{3}x + 6
    • (d) y=23x+1y = \frac{2}{3}x + 1

  • Step-by-Step Mathematical Derivation:

    • Extract the slope of the given reference line:
    • The reference equation is provided in slope-intercept form: y=23x+4y = \frac{2}{3}x + 4
    • Therefore, the slope of the given line is m=23m = \frac{2}{3}
    • Apply the parallel line condition:
    • Parallel lines must have identical slopes: mparallel=23m_{parallel} = \frac{2}{3}
    • Use the point-slope form with the target point (x1,y1)=(6,−1)(x_1, y_1) = (6, -1):
    • y−y1=m(x−x1)y - y_1 = m(x - x_1)
    • y−(−1)=23(x−6)y - (-1) = \frac{2}{3}(x - 6)
    • y+1=23x−(23)(6)y + 1 = \frac{2}{3}x - \left(\frac{2}{3}\right)(6)
    • y+1=23x−4y + 1 = \frac{2}{3}x - 4
    • Solve for yy to convert into slope-intercept form:
    • y=23x−4−1y = \frac{2}{3}x - 4 - 1
    • y=23x−5y = \frac{2}{3}x - 5

  • Verification of Distractors:

    • Option (b) uses m=−32m = -\frac{3}{2}, which is the perpendicular slope, not the parallel slope.
    • Option (c) uses an incorrect yy-intercept of +6+6, which would result from an addition error during isolation of yy.
    • Option (d) uses an incorrect yy-intercept of +1+1.
    • Correct Option: (a) y=23x−5y = \frac{2}{3}x - 5

Problem 8: Finding Linear Equations from Slope and X-Intercept

  • Problem Statement:

    • Determine the equation of the line that has slope of −2-2 and an xx-intercept of 32\frac{3}{2}.

  • Available Multiple-Choice Options:

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

  • Step-by-Step Mathematical Derivation:

    • Identify the given parameters:
    • Slope: m=−2m = -2
    • xx-intercept: a=32a = \frac{3}{2}, which specifies the coordinate point (32,0)(\frac{3}{2}, 0)
    • Apply point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1) using the intercept coordinates:
    • y−0=−2(x−32)y - 0 = -2\left(x - \frac{3}{2}\right)
    • Distribute the slope across the parentheses:
    • y=−2(x)−2(−32)y = -2(x) - 2\left(-\frac{3}{2}\right)
    • y=−2x+3y = -2x + 3

  • Verification of Distractors:

    • Option (a) incorrectly computes the constant as −3-3, which arises from a sign error when multiplying −2×−32-2 \times -\frac{3}{2}.
    • Options (c) and (d) erroneously use the reciprocal slope −12-\frac{1}{2} rather than the stated slope of −2-2.
    • Correct Option: (b) y=−2x+3y = -2x + 3

Problem 9: Slope Calculation Between Two Endpoints

  • Problem Statement:

    • A line segment has endpoints A(−7,5)A(-7, 5) and B(5,−1)B(5, -1). Determine the slope of ABAB.

  • Available Multiple-Choice Options:

    • (a) 22
    • (b) 33
    • (c) 11
    • (d) −12-\frac{1}{2}

  • Step-by-Step Mathematical Derivation:

    • Define the endpoint coordinates:
    • Point A=(x1,y1)=(−7,5)A = (x_1, y_1) = (-7, 5)
    • Point B=(x2,y2)=(5,−1)B = (x_2, y_2) = (5, -1)
    • Calculate the change in vertical coordinates (Δy\Delta y):
    • Δy=y2−y1=−1−5=−6\Delta y = y_2 - y_1 = -1 - 5 = -6
    • Calculate the change in horizontal coordinates (Δx\Delta x):
    • Δx=x2−x1=5−(−7)=5+7=12\Delta x = x_2 - x_1 = 5 - (-7) = 5 + 7 = 12
    • Compute the ratio of vertical change to horizontal change:
    • m=ΔyΔx=−612m = \frac{\Delta y}{\Delta x} = \frac{-6}{12}
    • Simplify the rational fraction to simplest form by dividing numerator and denominator by their greatest common divisor (66):
    • m=−12m = -\frac{1}{2}

  • Verification of Distractors:

    • If the run were mistakenly computed with subtraction as 5−7=−25 - 7 = -2, the calculation would yield −6−2=3\frac{-6}{-2} = 3, corresponding to distractor (b).
    • If the inverted ratio ΔxΔy\frac{\Delta x}{\Delta y} were computed, it would yield 12−6=−2\frac{12}{-6} = -2, corresponding to distractor (a).
    • Correct Option: (d) −12-\frac{1}{2}

Problem 10: Standard Slope-Intercept Form Identification

  • Problem Statement:

    • Which of the following equations has a graph with a slope of 23\frac{2}{3} and a yy-intercept of −45-\frac{4}{5}?

  • Available Multiple-Choice Options:

    • (a) y=23x−45y = \frac{2}{3}x - \frac{4}{5}
    • (b) y=23x+45y = \frac{2}{3}x + \frac{4}{5}
    • (c) y=−23x−45y = -\frac{2}{3}x - \frac{4}{5}
    • (d) y=32x+45y = \frac{3}{2}x + \frac{4}{5}

  • Step-by-Step Mathematical Derivation:

    • Identify the target slope and intercept values:
    • Slope: m=23m = \frac{2}{3}
    • yy-intercept: b=−45b = -\frac{4}{5}
    • Directly substitute into the standard slope-intercept form equation:
    • y=mx+by = mx + b
    • y=(23)x+(−45)y = \left(\frac{2}{3}\right)x + \left(-\frac{4}{5}\right)
    • y=23x−45y = \frac{2}{3}x - \frac{4}{5}

  • Verification of Distractors:

    • Option (b) has a positive yy-intercept of +45+\frac{4}{5}.
    • Option (c) has a negative slope of −23-\frac{2}{3}.
    • Option (d) has an inverted slope of 32\frac{3}{2} and a positive yy-intercept of +45+\frac{4}{5}.
    • Correct Option: (a) y=23x−45y = \frac{2}{3}x - \frac{4}{5}

Quick Reference Summary of Problem Solutions

  • Question 6 Solution Summary:

    • Slope of l1l_1: m=−43m = -\frac{4}{3}
    • Equation of l1l_1: y=−43x−283y = -\frac{4}{3}x - \frac{28}{3}
    • False statements: Statement I (yy-intercept is −283-\frac{28}{3}, not −7-7) and Statement II (l1l_1 passes through (8,−20)(8, -20), not (8,2)(8, 2)).
    • Correct Choice: (a)

  • Question 7 Solution Summary:

    • Parallel slope: m=23m = \frac{2}{3}
    • Passing point: (6,−1)(6, -1)
    • Equation: y=23x−5y = \frac{2}{3}x - 5
    • Correct Choice: (a)

  • Question 8 Solution Summary:

    • Given slope: m=−2m = -2
    • Given xx-intercept point: (32,0)(\frac{3}{2}, 0)
    • Equation: y=−2x+3y = -2x + 3
    • Correct Choice: (b)

  • Question 9 Solution Summary:

    • Endpoints: A(−7,5)A(-7, 5) and B(5,−1)B(5, -1)
    • Slope formula calculation: m=−1−55−(−7)=−612=−12m = \frac{-1 - 5}{5 - (-7)} = \frac{-6}{12} = -\frac{1}{2}
    • Correct Choice: (d)

  • Question 10 Solution Summary:

    • Given slope: m=23m = \frac{2}{3}
    • Given yy-intercept: b=−45b = -\frac{4}{5}
    • Slope-intercept equation: y=23x−45y = \frac{2}{3}x - \frac{4}{5}
    • Correct Choice: (a)