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Geometry and Trigonometry - Coordinate geometry: distance, midpoint, and gradient of a line

Grade 9IB

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

🔑Concepts

The Cartesian Plane: A 2D coordinate system defined by a horizontal xx-axis and a vertical yy-axis. Every point is represented as (x,y)(x, y), where xx is the horizontal distance from the origin (0,0)(0,0) and yy is the vertical distance. Visually, the plane is divided into four quadrants by these intersecting axes.

The Gradient (Slope): This measures the steepness and direction of a line. Visually, a line with a positive gradient goes 'up' from left to right, while a negative gradient goes 'down'. A gradient of zero indicates a horizontal line, and a vertical line has an undefined gradient. It is often described as the 'rise' (vertical change) over the 'run' (horizontal change).

Distance between Points: This is the straight-line length between two coordinates (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2). Conceptually, it is derived from the Pythagorean Theorem by treating the line segment as the hypotenuse of a right-angled triangle where the base is the change in xx and the height is the change in yy.

The Midpoint: The midpoint is the exact center of a line segment connecting two points. Visually, it is the point that bisects the line into two equal halves. To find it, you calculate the average of the xx-coordinates and the average of the yy-coordinates of the endpoints.

Collinear Points: Points that lie exactly on the same straight line are called collinear. To verify this visually or algebraically, the gradient between any two pairs of these points must be equal (e.g., the gradient of ABAB is the same as the gradient of BCBC).

Parallel and Perpendicular Lines: Lines that never meet are parallel and have identical gradients. Lines that meet at a 9090^{\circ} angle are perpendicular, and the product of their gradients is 1-1 (negative reciprocals), except in the case of vertical and horizontal lines.

The Linear Equation: A straight line can be represented by the formula y=mx+cy = mx + c, where mm is the gradient and cc is the yy-intercept. Visually, cc is the point where the line crosses the vertical yy-axis.

📐Formulae

Gradient: m=y2y1x2x1m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}

Distance: d=(x2x1)2+(y2y1)2d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}

Midpoint: M=(x1+x22,y1+y22)M = \left( \frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2} \right)

Equation of a line: y=mx+cy = mx + c

💡Examples

Problem 1:

Given two points A(2,3)A(-2, 3) and B(4,11)B(4, 11), calculate the gradient, the midpoint, and the distance between them.

Solution:

  1. Gradient (mm): m=1134(2)=86=43m = \frac{11 - 3}{4 - (-2)} = \frac{8}{6} = \frac{4}{3}. \n2. Midpoint (MM): M=(2+42,3+112)=(22,142)=(1,7)M = \left( \frac{-2 + 4}{2}, \frac{3 + 11}{2} \right) = \left( \frac{2}{2}, \frac{14}{2} \right) = (1, 7). \n3. Distance (dd): d=(4(2))2+(113)2=62+82=36+64=100=10d = \sqrt{(4 - (-2))^{2} + (11 - 3)^{2}} = \sqrt{6^{2} + 8^{2}} = \sqrt{36 + 64} = \sqrt{100} = 10.

Explanation:

To solve this, we identify x1=2,y1=3,x2=4,y2=11x_1 = -2, y_1 = 3, x_2 = 4, y_2 = 11 and substitute them into the standard formulas for coordinate geometry.

Problem 2:

The gradient of a line connecting P(1,4)P(1, 4) and Q(5,k)Q(5, k) is 22. Find the value of kk.

Solution:

  1. Use the gradient formula: m=y2y1x2x1m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}. \n2. Substitute known values: 2=k4512 = \frac{k - 4}{5 - 1}. \n3. Simplify the denominator: 2=k442 = \frac{k - 4}{4}. \n4. Multiply both sides by 4: 8=k48 = k - 4. \n5. Solve for kk: k=12k = 12.

Explanation:

This problem requires rearranging the gradient formula to solve for an unknown coordinate component when the slope is already known.