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Functions - Linear functions

Grade 11IB_AI

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

🔑Concepts

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The gradient (slope) mm of a linear function measures the rate of change. A positive gradient indicates the line rises from left to right, while a negative gradient indicates it falls. A zero gradient is a horizontal line, and an undefined gradient is a vertical line.

Graph showing a line with a positive gradient rising from left to right.
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The yy-intercept cc is the point where the line crosses the yy-axis, occurring at (0,c)(0, c). The xx-intercept is where the line crosses the xx-axis, occurring at (x,0)(x, 0), which is found by setting y=0y = 0 in the equation.

Graph identifying the x-intercept and y-intercept of a linear function.
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Parallel lines have identical gradients (m1=m2m_1 = m_2). This means they will never intersect and maintain a constant distance from each other.

Graph of two parallel lines with the same slope.
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Perpendicular lines intersect at a 90∘90^{\circ} angle. Their gradients are negative reciprocals of each other, satisfying the condition m1×m2=−1m_1 \times m_2 = -1.

Graph showing two perpendicular lines intersecting at the origin.

📐Formulae

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

y=mx+cy = mx + c

y−y1=m(x−x1)y - y_1 = m(x - x_1)

ax+by+d=0ax + by + d = 0

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

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

💡Examples

Problem 1:

Find the equation of the line passing through the points A(1,4)A(1, 4) and B(3,10)B(3, 10) in the form y=mx+cy = mx + c.

Solution:

  1. Calculate the gradient mm: m=10−43−1=62=3m = \frac{10 - 4}{3 - 1} = \frac{6}{2} = 3
  2. Use the point-gradient form with point A(1,4)A(1, 4): y−4=3(x−1)y - 4 = 3(x - 1)
  3. Expand and simplify to gradient-intercept form: y−4=3x−3y - 4 = 3x - 3 y=3x+1y = 3x + 1

Explanation:

First, the gradient is determined using the two-point formula. Then, one point is substituted into the point-gradient equation to find the final relationship between xx and yy.

Problem 2:

A line L1L_1 has the equation y=2x+5y = 2x + 5. Find the equation of line L2L_2 which is perpendicular to L1L_1 and passes through the point (4,−1)(4, -1).

Solution:

  1. Identify the gradient of L1L_1: m1=2m_1 = 2.
  2. Determine the perpendicular gradient m2m_2: m2=−1m1=−12m_2 = -\frac{1}{m_1} = -\frac{1}{2}
  3. Use the point-gradient form with point (4,−1)(4, -1): y−(−1)=−12(x−4)y - (-1) = -\frac{1}{2}(x - 4)
  4. Simplify: y+1=−12x+2y + 1 = -\frac{1}{2}x + 2 y=−12x+1y = -\frac{1}{2}x + 1

Explanation:

Perpendicular lines have gradients that are negative reciprocals. Once the new gradient is found, the equation is constructed using the given point.

Problem 3:

The cost of renting a car involves a fixed insurance fee plus a charge per kilometer driven. For 100100 km, the cost is 7070. For 250250 km, the cost is 130130. Find the linear model C(x)C(x) for the cost.

Solution:

  1. Let xx be the distance in km and CC be the cost. The points are (100,70)(100, 70) and (250,130)(250, 130).
  2. Find the gradient (cost per km): m=130−70250−100=60150=0.4m = \frac{130 - 70}{250 - 100} = \frac{60}{150} = 0.4
  3. Use C=mx+cC = mx + c with (100,70)(100, 70): 70=0.4(100)+c70 = 0.4(100) + c 70=40+c⇒c=3070 = 40 + c \Rightarrow c = 30
  4. The model is: C(x)=0.4x+30C(x) = 0.4x + 30

Explanation:

This is a real-world application where the gradient represents the variable rate (cost per km) and the yy-intercept represents the fixed cost (insurance fee).

Problem 4:

Determine the equation of the line LL that passes through the xx-intercept of 55 and the yy-intercept of −2-2. Express the answer in the form ax+by+d=0ax + by + d = 0.

Line passing through (5,0) and (0,-2).

Solution:

  1. Identify two points on the line: A(5,0)A(5, 0) and B(0,−2)B(0, -2).
  2. Calculate the gradient mm: m=−2−00−5=−2−5=25m = \frac{-2 - 0}{0 - 5} = \frac{-2}{-5} = \frac{2}{5}
  3. Use the yy-intercept form y=mx+cy = mx + c: y=25x−2y = \frac{2}{5}x - 2
  4. Convert to general form ax+by+d=0ax + by + d = 0 by multiplying by 55: 5y=2x−105y = 2x - 10 2x−5y−10=02x - 5y - 10 = 0

Explanation:

To find the linear equation from intercepts, we identify the coordinates (5,0)(5,0) and (0,−2)(0,-2), compute the slope, and then rearrange the equation into the standard form required by the problem.

Problem 5:

A straight line LL passes through the point P(2,6)P(2, 6) and has a gradient of m=−1.5m = -1.5. Find the xx-intercept of this line. Illustrate the line on a coordinate grid.

Graph of the linear function y = -1.5x + 9 showing point P(2,6) and the x-intercept at (6,0).

Solution:

Step 1: Use the point-gradient formula y−y1=m(x−x1)y - y_1 = m(x - x_1). y−6=−1.5(x−2)y - 6 = -1.5(x - 2)

Step 2: Simplify the equation to the form y=mx+cy = mx + c. y−6=−1.5x+3y - 6 = -1.5x + 3 y=−1.5x+9y = -1.5x + 9

Step 3: To find the xx-intercept, set y=0y = 0. 0=−1.5x+90 = -1.5x + 9 1.5x=91.5x = 9 x=91.5=6x = \frac{9}{1.5} = 6

The xx-intercept is (6,0)(6, 0).

Explanation:

The point-gradient form is the most efficient way to find the equation when a point and the slope are known. The xx-intercept is the point where the line crosses the horizontal axis, which always occurs when the yy-coordinate is zero.