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Geometry and Trigonometry - Coordinate geometry (distance, midpoint, gradient)

Grade 10IB

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

🔑Concepts

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The Cartesian Plane: A 2D coordinate system defined by a horizontal x-axis and a vertical y-axis that intersect at the origin (0,0)(0,0). Any point PP is described by an ordered pair (x,y)(x, y), representing its horizontal and vertical displacement from the origin.

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Distance between Two Points: This represents the length of the straight line segment connecting points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2). Visually, this creates a right-angled triangle where the distance is the hypotenuse, and the legs are the horizontal change (x2−x1)(x_2 - x_1) and vertical change (y2−y1)(y_2 - y_1).

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Midpoint of a Segment: The midpoint is the exact center point of a line segment. Conceptually, it is the 'average' of the coordinates; visually, it lies halfway along the line connecting the two endpoints, dividing the segment into two equal lengths.

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Gradient (Slope): The gradient mm measures the steepness and direction of a line. It is defined as the 'rise over run'. Visually, a positive gradient slopes upwards from left to right, a negative gradient slopes downwards, a horizontal line has a gradient of 00, and a vertical line has an undefined gradient.

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Parallel Lines: Two lines are parallel if they have the same gradient (m1=m2m_1 = m_2). Visually, these lines move in the same direction and will never intersect, maintaining a constant distance between each other regardless of how far they are extended.

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Perpendicular Lines: Two lines are perpendicular if they intersect at a right angle (90∘90^\circ). Their gradients are negative reciprocals of each other, meaning m1×m2=−1m_1 \times m_2 = -1. Visually, one line is rotated 90∘90^\circ relative to the other.

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Collinear Points: Three or more points are said to be collinear if they all lie on the same straight line. This can be verified by checking if the gradient between any two pairs of these points is identical.

📐Formulae

Distance Formula: d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Midpoint Formula: M=(x1+x22,y1+y22)M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)

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

Gradient-Intercept form of a line: y=mx+cy = mx + c

Point-Gradient form of a line: y−y1=m(x−x1)y - y_1 = m(x - x_1)

Parallel Lines Condition: m1=m2m_1 = m_2

Perpendicular Lines Condition: m1×m2=−1m_1 \times m_2 = -1 or m2=−1m1m_2 = -\frac{1}{m_1}

💡Examples

Problem 1:

Given two points A(2,−3)A(2, -3) and B(6,5)B(6, 5), calculate the distance between them and the coordinates of the midpoint of segment ABAB.

Solution:

  1. To find the distance: d=(6−2)2+(5−(−3))2=42+82=16+64=80≈8.94d = \sqrt{(6 - 2)^2 + (5 - (-3))^2} = \sqrt{4^2 + 8^2} = \sqrt{16 + 64} = \sqrt{80} \approx 8.94.
  2. To find the midpoint: M=(2+62,−3+52)=(82,22)=(4,1)M = \left( \frac{2 + 6}{2}, \frac{-3 + 5}{2} \right) = \left( \frac{8}{2}, \frac{2}{2} \right) = (4, 1).

Explanation:

We use the distance formula derived from Pythagoras' theorem and the midpoint formula to find the average of the xx and yy coordinates.

Problem 2:

Find the equation of the line that passes through the point (4,2)(4, 2) and is perpendicular to a line with gradient m=−2m = -2.

Solution:

  1. Identify the gradient of the perpendicular line: m⊥=−1−2=12m_{\perp} = -\frac{1}{-2} = \frac{1}{2}.
  2. Use the point-gradient formula with point (4,2)(4, 2) and m=12m = \frac{1}{2}: y−2=12(x−4)y - 2 = \frac{1}{2}(x - 4) y−2=12x−2y - 2 = \frac{1}{2}x - 2 y=12xy = \frac{1}{2}x.

Explanation:

First, we find the perpendicular gradient by taking the negative reciprocal. Then, we substitute the gradient and the given coordinates into the point-gradient equation to solve for the final linear equation.