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Geometry and Trigonometry - Equation of a straight line

Grade 10IB

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

🔑Concepts

The Gradient (mm) measures the steepness and direction of a line. Visually, a positive gradient slopes upwards from left to right, while a negative gradient slopes downwards. A horizontal line has a gradient of 00, and a vertical line has an undefined gradient.

The yy-intercept (cc) is the point where the line crosses the vertical yy-axis. On a graph, this is located at the coordinates (0,c)(0, c). Conversely, the xx-intercept is where the line crosses the horizontal xx-axis, found by setting y=0y = 0.

The Gradient-Intercept form, y=mx+cy = mx + c, is the most common way to express a straight line. Here, mm represents the 'rise over run' (vertical change divided by horizontal change) and cc represents the height at which the line hits the yy-axis.

The Point-Gradient form, yy1=m(xx1)y - y_1 = m(x - x_1), is used to find the equation of a line when you know its slope and one specific point (x1,y1)(x_1, y_1) that it passes through. This is often rearranged into other forms for the final answer.

Parallel lines are lines that maintain a constant distance from each other and never meet. Visually, they have the same slant, which means their gradients are equal (m1=m2m_1 = m_2).

Perpendicular lines meet at a right angle (9090^{\circ}). Their gradients are negative reciprocals of each other (m1×m2=1m_1 \times m_2 = -1). For example, if one line has a gradient of 23\frac{2}{3}, the perpendicular line has a gradient of 32-\frac{3}{2}.

The Midpoint is the exact center point between two coordinates (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2). Visually, it is the 'average' position of the two points on the Cartesian plane.

Distance between two points represents the length of the straight-line segment connecting them. It is calculated by visualizing a right-angled triangle between the points and applying the Pythagorean theorem to the horizontal and vertical distances.

📐Formulae

Gradient: m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}

Gradient-Intercept Form: y=mx+cy = mx + c

Point-Gradient Form: yy1=m(xx1)y - y_1 = m(x - x_1)

General Form: ax+by+d=0ax + by + d = 0 (where a,b,da, b, d are usually integers)

Parallel Condition: m1=m2m_1 = m_2

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

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

Distance Formula: d=(x2x1)2+(y2y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

💡Examples

Problem 1:

Find the equation of the line passing through the points A(2,5)A(2, 5) and B(4,9)B(4, 9). Give your answer in the form y=mx+cy = mx + c.

Solution:

Step 1: Calculate the gradient (mm): m=y2y1x2x1=9542=42=2m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{9 - 5}{4 - 2} = \frac{4}{2} = 2

Step 2: Use the point-gradient formula with point A(2,5)A(2, 5): y5=2(x2)y - 5 = 2(x - 2)

Step 3: Expand and simplify: y5=2x4y - 5 = 2x - 4 y=2x+1y = 2x + 1

Explanation:

To find the equation of a line from two points, first determine the rate of change (gradient) between them. Then, substitute one point and the gradient into the linear equation model to solve for the intercept.

Problem 2:

Line L1L_1 has the equation y=3x4y = 3x - 4. Find the equation of line L2L_2 which is perpendicular to L1L_1 and passes through the point (6,1)(6, 1).

Solution:

Step 1: Identify the gradient of L1L_1. m1=3m_1 = 3.

Step 2: Determine the perpendicular gradient (m2m_2): m2=1m1=13m_2 = -\frac{1}{m_1} = -\frac{1}{3}

Step 3: Use the point (6,1)(6, 1) and m2=13m_2 = -\frac{1}{3} in the point-gradient form: y1=13(x6)y - 1 = -\frac{1}{3}(x - 6)

Step 4: Distribute and simplify: y1=13x+2y - 1 = -\frac{1}{3}x + 2 y=13x+3y = -\frac{1}{3}x + 3

Explanation:

Since the lines are perpendicular, their gradients must be negative reciprocals. After finding the new gradient, the point-slope formula is used to derive the specific equation for the second line.