Lesson 7: Linear Equations and Intersections
Learning Objectives
By the end of this lesson, students will be able to:
- Write equations of lines in slope-intercept and point-slope forms
- Convert between equation forms
- Find the intersection point of two lines
Equations of Lines
Slope-Intercept Form
\[y = mx + c\]
Point-Slope Form
\[y - y_1 = m(x - x_1)\]
General Form
\[ax + by + c = 0\]
Example 1: Equation from Gradient and Point
Write the equation of the line with gradient 3 passing through (2, 5).
Solution: \[y - 5 = 3(x - 2)\] \[y - 5 = 3x - 6\] \[y = 3x - 1\]
Example 2: Equation from Two Points
Find the equation of the line through P(1, 3) and Q(4, 9).
Solution: \[m = \frac{9 - 3}{4 - 1} = 2\] \[y - 3 = 2(x - 1)\] \[y = 2x + 1\]
Intersection of Two Lines
The intersection point solves both equations at the same time.
Example 3: Intersection by Substitution
Find the intersection of:
\[y = 2x + 1\] \[y = -x + 7\]
Solution: Set them equal: \[2x + 1 = -x + 7\] \[3x = 6\] \[x = 2\] Substitute back: \[y = 2(2) + 1 = 5\]
Intersection is \((2, 5)\).
Example 4: Intersection with General Form
Find the intersection of:
\[2x + y - 8 = 0\] \[x - y + 1 = 0\]
Solution: From the second equation: \(y = x + 1\). Substitute into the first: \[2x + (x + 1) - 8 = 0\] \[3x - 7 = 0\] \[x = \frac{7}{3}\] Then \(y = \frac{7}{3} + 1 = \frac{10}{3}\).
Intersection is \(\left(\frac{7}{3}, \frac{10}{3}\right)\).
Practice Exercises
- Write the equation of the line:
- With gradient -2 and y-intercept 5
- Through point (3, 7) with gradient -1
- Through points (0, 2) and (5, 12)
- Convert to slope-intercept form:
- \(3x + y - 6 = 0\)
- \(2x - 4y + 8 = 0\)
- Find the intersection point of each pair:
- \(y = x + 3\) and \(y = -2x + 9\)
- \(y = 4x - 5\) and \(2x + y = 7\)
Worksheet 2B: Linear Equations and Intersections
- Write the equation of the line:
- With gradient 3 and y-intercept 5
- With gradient -2 passing through (4, 1)
- Through points (2, 3) and (6, 11)
- Convert to slope-intercept form (\(y = mx + c\)):
- \(x + 2y - 10 = 0\)
- \(4x - 2y + 6 = 0\)
- Find the intersection point:
- \(y = 2x - 1\) and \(y = -x + 8\)
- \(3x + y = 12\) and \(x - y = 4\)
- \(y = 3x + 5\), \(y = -2x + 9\), \(y = 2x - 1\)
- \(y = -\frac{1}{2}x + 5\), \(y = 2x + 3\)
- (3, 5), (4, 0)
Previous Lesson: Lesson 6: Gradient Using the Formula
Next Lesson: Lesson 8: Parallel Lines