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Coordinate Geometry - The Concept of Slope (Gradient)-advanced

Grade 9CBSE

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

🔑Concepts

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The slope mm of a line is a measure of its steepness and direction. It is defined as the ratio of the vertical change ('rise') to the horizontal change ('run') between any two points on the line. For a line passing through (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2), the slope is m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}.

A line showing the geometric interpretation of rise over run between two points A and B.
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The slope of a line is also equal to the tangent of the angle θ\theta that the line makes with the positive direction of the xx-axis (measured counter-clockwise). Thus, m=tan⁡(θ)m = \tan(\theta). If the line is horizontal, θ=0∘\theta = 0^{\circ} and m=0m = 0. If the line is vertical, θ=90∘\theta = 90^{\circ} and the slope is undefined.

Line intersecting the x-axis at an angle theta.
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Two non-vertical lines are parallel if and only if their slopes are equal (m1=m2m_1 = m_2). They are perpendicular if and only if the product of their slopes is −1-1 (m1×m2=−1m_1 \times m_2 = -1).

Visual representation of parallel and perpendicular lines with slope conditions.
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Three points A,BA, B, and CC are collinear if the slope of segment ABAB is equal to the slope of segment BCBC. This implies all three points lie on the same straight line.

Three collinear points A, B, and C on a single line.

📐Formulae

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

m=tan⁡(θ)m = \tan(\theta) indices where 0≤θ<180∘0 \le \theta < 180^{\circ}

m1=m2 (Condition for Parallel Lines)m_1 = m_2 \text{ (Condition for Parallel Lines)}

m1×m2=−1 (Condition for Perpendicular Lines)m_1 \times m_2 = -1 \text{ (Condition for Perpendicular Lines)}

Slope of AB=Slope of BC  ⟹  A,B,C are collinear\text{Slope of AB} = \text{Slope of BC} \implies A, B, C \text{ are collinear}

💡Examples

Problem 1:

Find the slope of a line passing through the points P(−2,3)P(-2, 3) and Q(4,−5)Q(4, -5).

Solution:

Given points: (x1,y1)=(−2,3)(x_1, y_1) = (-2, 3) and (x2,y2)=(4,−5)(x_2, y_2) = (4, -5). Using the slope formula: m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1} m=−5−34−(−2)m = \frac{-5 - 3}{4 - (-2)} m=−84+2m = \frac{-8}{4 + 2} m=−86m = \frac{-8}{6} m=−43m = -\frac{4}{3} The slope is −43-\frac{4}{3}.

Explanation:

We apply the standard coordinate slope formula by substituting the given coordinates, ensuring the subtraction of negative numbers is handled correctly.

Problem 2:

If the line joining A(k,2)A(k, 2) and B(3,4)B(3, 4) is perpendicular to the line joining C(1,2)C(1, 2) and D(2,5)D(2, 5), find the value of kk.

Solution:

First, find the slope of line CDCD (m1m_1): m1=5−22−1=31=3m_1 = \frac{5 - 2}{2 - 1} = \frac{3}{1} = 3 Next, find the slope of line ABAB (m2m_2): m2=4−23−k=23−km_2 = \frac{4 - 2}{3 - k} = \frac{2}{3 - k} Since the lines are perpendicular, m1×m2=−1m_1 \times m_2 = -1: 3×(23−k)=−13 \times \left(\frac{2}{3 - k}\right) = -1 63−k=−1\frac{6}{3 - k} = -1 6=−1(3−k)6 = -1(3 - k) 6=−3+k6 = -3 + k k=6+3=9k = 6 + 3 = 9 The value of kk is 99.

Explanation:

For perpendicular lines, the product of slopes is −1-1. We calculate both slopes and solve the resulting algebraic equation for the unknown variable kk.

Problem 3:

Show that the points A(1,1)A(1, 1), B(2,3)B(2, 3), and C(3,5)C(3, 5) are collinear using the concept of slope.

Solution:

Calculate the slope of ABAB (mABm_{AB}): mAB=3−12−1=21=2m_{AB} = \frac{3 - 1}{2 - 1} = \frac{2}{1} = 2 Calculate the slope of BCBC (mBCm_{BC}): mBC=5−33−2=21=2m_{BC} = \frac{5 - 3}{3 - 2} = \frac{2}{1} = 2 Since mAB=mBC=2m_{AB} = m_{BC} = 2, and BB is a common point, the points A,B,A, B, and CC lie on the same straight line.

Explanation:

Collinearity is proven when the slope between consecutive pairs of points is the same, indicating they follow the same path.

Problem 4:

A line passes through the point A(1,2)A(1, 2) and is parallel to the line joining P(4,5)P(4, 5) and Q(−2,1)Q(-2, 1). Find the slope of the line and its inclination θ\theta.

Two parallel lines in a coordinate plane, one passing through A and the other through P and Q.

Solution:

  1. First, find the slope of the line PQPQ: mPQ=1−5−2−4=−4−6=23m_{PQ} = \frac{1 - 5}{-2 - 4} = \frac{-4}{-6} = \frac{2}{3}
  2. Since the required line is parallel to PQPQ, its slope mm is the same: m=23m = \frac{2}{3}
  3. To find the inclination θ\theta: tan⁡(θ)=23\tan(\theta) = \frac{2}{3} θ=tan⁡−1(23)≈33.69∘\theta = \tan^{-1}\left(\frac{2}{3}\right) \approx 33.69^{\circ}

Explanation:

Parallel lines have identical slopes. We calculate the slope of the known line segment and apply that value to the line passing through point A.

Problem 5:

Find the value of xx such that the line joining the points A(x,−1)A(x, -1) and B(2,1)B(2, 1) is perpendicular to the line joining C(4,5)C(4, 5) and D(0,3)D(0, 3). Represent these lines graphically.

A coordinate plane showing two perpendicular lines. Line AB passes through (3, -1) and (2, 1). Line CD passes through (4, 5) and (0, 3).

Solution:

  1. First, calculate the slope (m1m_1) of line ABAB: m1=1−(−1)2−x=22−xm_1 = \frac{1 - (-1)}{2 - x} = \frac{2}{2 - x}

  2. Next, calculate the slope (m2m_2) of line CDCD: m2=3−50−4=−2−4=12m_2 = \frac{3 - 5}{0 - 4} = \frac{-2}{-4} = \frac{1}{2}

  3. Since line ABAB is perpendicular to line CDCD, the product of their slopes must be −1-1: m1×m2=−1m_1 \times m_2 = -1 (22−x)×(12)=−1\left( \frac{2}{2 - x} \right) \times \left( \frac{1}{2} \right) = -1

  4. Simplify the equation: 12−x=−1\frac{1}{2 - x} = -1 1=−(2−x)1 = -(2 - x) 1=−2+x1 = -2 + x x=3x = 3

Thus, the value of xx is 33.

Explanation:

The concept of perpendicularity in coordinate geometry states that if two lines are perpendicular, the product of their slopes is −1-1 (m1×m2=−1m_1 \times m_2 = -1). We find the expressions for the slopes of both lines using the formula m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1} and solve for the unknown coordinate.