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Coordinate Geometry - Properties of Slope-advanced

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The slope (or gradient) mm of a line measures its steepness and direction. For a line passing through (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2), the slope is the ratio of 'rise' to 'run'. A line rising to the right has a positive slope (m>0m > 0), while a line falling to the right has a negative slope (m<0m < 0).

A coordinate plane showing a line segment AB with rise and run components labeled to illustrate slope calculation.
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Parallel lines have identical slopes (m1=m2m_1 = m_2) because they maintain the same angle of inclination with the positive xx-axis. Conversely, if two lines have the same slope, they must be parallel (or coincident).

Two parallel lines on a coordinate grid showing the same slope.
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The slopes of perpendicular lines are negative reciprocals of each other, meaning m1×m2=−1m_1 \times m_2 = -1. This property allows us to find the slope of a line that meets another at a 90∘90^{\circ} angle.

Two lines intersecting at a right angle with slopes m and -1/m.
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The slope of a horizontal line (xx-axis or parallel to it) is always 00 because the change in yy is zero. The slope of a vertical line (yy-axis or parallel to it) is undefined because the change in xx is zero, leading to division by zero.

📐Formulae

m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}

Parallel lines: m1=m2\text{Parallel lines: } m_1 = m_2

Perpendicular lines: m1×m2=−1\text{Perpendicular lines: } m_1 \times m_2 = -1

Condition for Collinearity: y2−y1x2−x1=y3−y2x3−x2\text{Condition for Collinearity: } \frac{y_2 - y_1}{x_2 - x_1} = \frac{y_3 - y_2}{x_3 - x_2}

💡Examples

Problem 1:

Find the slope of the line passing through the points P(2,−3)P(2, -3) and Q(5,6)Q(5, 6).

Solution:

Let (x1,y1)=(2,−3)(x_1, y_1) = (2, -3) and (x2,y2)=(5,6)(x_2, y_2) = (5, 6). Using the slope formula: m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1} m=6−(−3)5−2m = \frac{6 - (-3)}{5 - 2} m=6+33m = \frac{6 + 3}{3} m=93=3m = \frac{9}{3} = 3 So, the slope of the line is 33.

Explanation:

We apply the standard slope formula by substituting the given coordinates and simplifying the fraction.

Problem 2:

Check if the points A(−1,1)A(-1, 1), B(1,3)B(1, 3), and C(3,5)C(3, 5) are collinear using the property of slope.

Solution:

First, find the slope of ABAB (m1m_1): m1=3−11−(−1)=22=1m_1 = \frac{3 - 1}{1 - (-1)} = \frac{2}{2} = 1 Next, find the slope of BCBC (m2m_2): m2=5−33−1=22=1m_2 = \frac{5 - 3}{3 - 1} = \frac{2}{2} = 1 Since m1=m2m_1 = m_2, the slopes are equal.

Explanation:

If the slope of ABAB equals the slope of BCBC, then the segments have the same direction and share point BB, meaning A,B,A, B, and CC lie on the same line.

Problem 3:

A line L1L_1 passes through (1,2)(1, 2) and (3,8)(3, 8). Another line L2L_2 is perpendicular to L1L_1. Find the slope of L2L_2.

Solution:

First, calculate the slope of L1L_1 (m1m_1): m1=8−23−1=62=3m_1 = \frac{8 - 2}{3 - 1} = \frac{6}{2} = 3 Since L2⊥L1L_2 \perp L_1, the product of their slopes must be −1-1: m1×m2=−1m_1 \times m_2 = -1 3×m2=−13 \times m_2 = -1 m2=−13m_2 = -\frac{1}{3} Thus, the slope of L2L_2 is −13-\frac{1}{3}.

Explanation:

The product of slopes of perpendicular lines is −1-1. We find the first slope and then take its negative reciprocal.

Problem 4:

Line L1L_1 passes through the points A(2,3)A(2, 3) and B(4,k)B(4, k). If line L1L_1 is parallel to line L2L_2 which has a slope of 22, find the value of kk.

A line passing through (2,3) and (4,k) with a slope of 2.

Solution:

  1. Since L1L_1 is parallel to L2L_2, their slopes must be equal: m1=m2=2m_1 = m_2 = 2.
  2. Use the slope formula for points A(2,3)A(2, 3) and B(4,k)B(4, k): m1=k−34−2m_1 = \frac{k - 3}{4 - 2}
  3. Substitute the known slope: 2=k−322 = \frac{k - 3}{2}
  4. Solve for kk: 4=k−34 = k - 3 k=7k = 7

Explanation:

Parallel lines share the same slope. By setting the slope calculated from the coordinates equal to the given slope of the parallel line, we can solve for the unknown coordinate.

Problem 5:

Determine the value of xx such that the line through (x,4)(x, 4) and (2,1)(2, 1) is perpendicular to the line passing through (3,5)(3, 5) and (5,9)(5, 9).

Two perpendicular lines L1 and L2 on a coordinate system.

Solution:

  1. Let m1m_1 be the slope of the line through (x,4)(x, 4) and (2,1)(2, 1): m1=1−42−x=−32−xm_1 = \frac{1 - 4}{2 - x} = \frac{-3}{2 - x}
  2. Let m2m_2 be the slope of the line through (3,5)(3, 5) and (5,9)(5, 9): m2=9−55−3=42=2m_2 = \frac{9 - 5}{5 - 3} = \frac{4}{2} = 2
  3. Since the lines are perpendicular, m1×m2=−1m_1 \times m_2 = -1: (−32−x)×2=−1\left(\frac{-3}{2 - x}\right) \times 2 = -1 −62−x=−1\frac{-6}{2 - x} = -1
  4. Solve for xx: −6=−1(2−x)-6 = -1(2 - x) −6=−2+x-6 = -2 + x x=−4x = -4

Explanation:

First, find the slope of the second line. Use the perpendicularity condition (product of slopes is -1) to determine the required slope for the first line, then solve for the unknown coordinate.