Tenth Grade Mathematics: Slope of a Line, Parallel, and Perpendicular Conditions
Every straight line in the coordinate plane possesses a definite orientation and steepness relative to the horizontal axis. In coordinate geometry, this directional steepness is quantified as the slope (or gradient) of the line. Derived as the trigonometric tangent of the angle of inclination, the slope provides a direct algebraic test to determine whether two lines run strictly parallel, cross obliquely, or meet at right angles. Understanding slopes unites coordinate geometry with trigonometry and prepares students for the study of derivatives.
The Angle of Inclination and Definition of Slope
The inclination theta of a non-horizontal straight line is the angle that the line makes with the positive direction of the x-axis, measured in the counterclockwise direction:
Range of Inclination:
0 <= theta < 180 degrees
Visualizing Angle of Inclination:
y
| / Line L
| /
| /
| / theta (Measured counterclockwise
| / from positive x-axis)
-------------------+-----+------------------- x
^
Formal Definition of Slope:
If theta is the inclination of a line L (with theta != 90 degrees), then the
slope m of the line is defined as the tangent of theta:
m = tan(theta)
Behavior of Slope based on Inclination:
1. Acute Angle (0 < theta < 90 deg): Slope is positive (m > 0). Line rises to right.
2. Obtuse Angle (90 < theta < 180 deg): Slope is negative (m < 0). Line falls to right.
3. Horizontal Line (theta = 0 deg): Slope is zero (m = tan(0) = 0).
4. Vertical Line (theta = 90 deg): Slope is undefined (m = tan(90) = undefined).
Slope Formula Through Two Given Points
Let P(x1, y1) and Q(x2, y2) be two distinct points on a non-vertical line (where x1 != x2):
Theorem (Two-Point Slope Formula):
The slope m of the line passing through (x1, y1) and (x2, y2) is:
y2 - y1 Vertical Rise
m = --------- = ---------------
x2 - x1 Horizontal Run
Proof: In right-angled triangle PQR constructed between the points, the angle at P equals the inclination theta (corresponding angles). By trigonometry: tan(theta) = Opposite / Adjacent = (y2 - y1) / (x2 - x1).
Conditions for Parallel and Perpendicular Lines
Let L1 and L2 be two non-vertical lines having slopes m1 and m2 respectively:
1. Condition for Parallel Lines:
Two lines are parallel if and only if their inclinations are equal (theta1 = theta2):
m1 = m2
Parallel lines have IDENTICAL slopes.
2. Condition for Perpendicular Lines:
Two lines are perpendicular (intersecting at 90 degrees) if and only if:
theta2 = theta1 + 90 deg ===> tan(theta2) = tan(theta1 + 90) = -cot(theta1) = -1 / tan(theta1)
m2 = -1 / m1
Multiplying both sides by m1:
m1 * m2 = -1
The product of the slopes of two perpendicular lines is -1 (negative reciprocals).
Summary Table:
+---------------------+----------------------------+-----------------------+
| Geometric Relation | Condition on Slopes | Algebraic Expression |
+---------------------+----------------------------+-----------------------+
| Parallel Lines | Slopes are equal | m1 = m2 |
+---------------------+----------------------------+-----------------------+
| Perpendicular Lines | Slopes are negative | m1 * m2 = -1 |
| | reciprocals | (or m2 = -1/m1) |
+---------------------+----------------------------+-----------------------+
Standard Forms of the Equation of a Straight Line
1. Slope-Intercept Form:
y = m*x + c (m = slope, c = y-intercept (0, c))
2. Point-Slope Form:
y - y1 = m * (x - x1) (passes through (x1, y1) with slope m)
3. Two-Point Form:
y - y1 = [(y2 - y1)/(x2 - x1)] * (x - x1)
4. General Linear Form:
A*x + B*y + C = 0 ===> Slope m = -A / B
Chapter Exercises
Exercise 1: Find the slope of the line: (a) passing through the points (3, -2) and (-1, 4), and (b) making an angle of 60 degrees with the positive direction of the x-axis.
Exercise 2: Line L1 passes through the points (2, 5) and (-3, 6). Line L2 passes through (-3, 5) and (k, 1). Find the value of k if: (a) L1 is parallel to L2, and (b) L1 is perpendicular to L2.
Exercise 3: Find the equation of the line that passes through the point (2, -3) and is perpendicular to the line joining (4, 1) and (-2, 3).
Solutions and Explanations
Solution 1: For part (a), the points are (x1, y1) = (3, -2) and (x2, y2) = (-1, 4). The slope formula gives: m = (y2 - y1)/(x2 - x1) = (4 - (-2))/(-1 - 3) = (4 + 2)/(-4) = 6/(-4) = -3/2. For part (b), the angle of inclination is theta = 60 degrees. The slope is m = tan(60 deg) = sqrt(3).
Solution 2: First find the slope of line L1 passing through (2, 5) and (-3, 6): m1 = (6 - 5)/(-3 - 2) = 1/(-5) = -1/5. Next find the slope of line L2 passing through (-3, 5) and (k, 1): m2 = (1 - 5)/(k - (-3)) = -4/(k + 3). For part (a), if L1 is parallel to L2, their slopes are equal: m1 = m2 ==> -1/5 = -4/(k + 3). Cross-multiplying: -(k + 3) = -20 ==> k + 3 = 20 ==> k = 17. For part (b), if L1 is perpendicular to L2, m1 * m2 = -1 ==> (-1/5) * [-4/(k + 3)] = -1 ==> 4 / [5(k + 3)] = -1 ==> 4 = -5(k + 3) ==> 4 = -5k - 15 ==> 5k = -19 ==> k = -19/5.
Solution 3: First compute the slope of the given line segment joining (4, 1) and (-2, 3): m_given = (3 - 1)/(-2 - 4) = 2/(-6) = -1/3. Since the required line is perpendicular to this line, its slope m must satisfy: m * (-1/3) = -1 ==> m = 3. The required line passes through the point (x1, y1) = (2, -3) with slope m = 3. Using point-slope form: y - y1 = m(x - x1) ==> y - (-3) = 3(x - 2) ==> y + 3 = 3x - 6 ==> 3x - y - 9 = 0. Thus, the equation of the line is 3*x - y - 9 = 0.