TutorChase logo
CIE A-Level Maths Study Notes

1.3.2 Forms of a Linear Equation

Linear equations, a cornerstone in algebra, offer a gateway to understanding complex mathematical concepts. This comprehensive guide explores the various forms of linear equations, their practical applications, and techniques for interconversion.

1. Slope-Intercept Form (y=mx+c)(y = mx + c)

The slope-intercept form, represented as y=mx+cy = mx + c, is invaluable for quickly identifying the slope (m)(m) and y-intercept (c)(c) of a line. It's particularly useful in graphing linear equations and understanding their behavior.

Slope-Intercept Form

Image courtesy of Cuemath

Example:

Suppose we want to find the equation of a line with a slope of 4 and passing through the point (1, 3).

Solution:

Given:

  • Slope (m): 44
  • Point: (1,3)(1, 3)

1. Use the Slope-Intercept Form

The slope-intercept form of a line is: y=mx+cy = mx + c

Where mm is the slope and cc is the y-intercept.

2. Substitute the Given Point

Substitute x=1x = 1 and y=3y = 3 into the equation to find cc:

3=4×1+c3 = 4 \times 1 + c

3: Solve for c

Solve the equation for cc:

3=4+c3 = 4 + c

c=34c = 3 - 4

c=1c = -1

With m=4m = 4 and c=1c = -1, the equation of the line is: y=4x1y = 4x - 1

This is the equation of the line with a slope of 44 passing through the point (1,3)(1, 3). It shows the relationship between xx and yy for every point on this line.

2. Point-Slope Form (yy1=m(xx1))(y - y_1 = m(x - x_1))

The point-slope form, yy1=m(xx1)y - y_1 = m(x - x_1), is essential when the slope and a specific point on the line are known. It's particularly useful in problems involving tangents to curves or constructing lines parallel or perpendicular to a given line.

Point slope form


Image courtesy of Geeksforgeeks

Example:

Find the equation of a line that is perpendicular to y=2x5y = 2x - 5 and passes through (3, 4).

Solution:

1. Determine the Slope of the Perpendicular Line

  • The given line is y=2x5y = 2x - 5.
  • The slope of this line is the coefficient of xx, which is 22.
  • The slope of a line perpendicular to this would be the negative reciprocal of 22. The negative reciprocal of mm is 1m-\frac{1}{m}.
  • Therefore, the slope of the perpendicular line is 12-\frac{1}{2}.

2. Use the Point-Slope Formula

  • The point-slope formula is yy1=m(xx1)y - y_1 = m(x - x_1), where ( m ) is the slope and (x1,y1)(x_1, y_1) is the point through which the line passes.
  • In this case, m=12m = -\frac{1}{2} and the point is (3,4)(3, 4).
  • Substituting these values, the result is: y4=12(x3)y - 4 = -\frac{1}{2}(x - 3)

3. Simplify the Equation

y=mx+by = mx + b

=y4=12x+32=y - 4 = -\frac{1}{2}x + \frac{3}{2}

=y=12x+32+4=y = -\frac{1}{2}x + \frac{3}{2} + 4

=y=12x+112=y = -\frac{1}{2}x + \frac{11}{2}

The equation of the line that is perpendicular to y=2x5y = 2x - 5 and passes through the point (3,4)(3, 4) is: y=12x+112y = -\frac{1}{2}x + \frac{11}{2}

3. General Form (ax+by+c=0)(ax + by + c = 0)

The general form, ax+by+c=0ax + by + c = 0, is a versatile representation used in various mathematical disciplines. It's particularly useful in systems of linear equations and in computational geometry.

general form of the line


Example:

Convert the equation 3y=6x93y = 6x - 9 into the general form.

Solution:

1. Start with the Given Equation

The given equation is: 3y=6x93y = 6x - 9

2. Rearrange the Equation

Rearrange this equation to get it into the form ax+by+c=0ax + by + c = 0.

6x+3y=9-6x + 3y = -9

3. Adjust the Equation to Match the General Form

Add ( 9 ) to both sides:

6x+3y+9=0-6x + 3y + 9 = 0

The equation 3y=6x93y = 6x - 9 in the general form is: 6x+3y+9=0-6x + 3y + 9 = 0

Intersection Points

Finding intersection points of lines is crucial in understanding their relationships and is a key concept in algebra and geometry.

intersection points

Image courtesy of Cuemath

Example:

Determine the intersection point of the lines y=x+6y = -x + 6 and y=2x+4y = 2x + 4.

Solution:

1. Set the Equations Equal to Each Other

x+6=2x+4 -x + 6 = 2x + 4

2. Solve for x

Solve this equation for x+6=2x+4 -x + 6 = 2x + 4 :

x2x=46-x - 2x = 4 - 6 3x=2-3x = -2 x=23x = \frac{-2}{-3} x=23x = \frac{2}{3}

3. Solve for y

Substitute x=23x = \frac{2}{3} to find yy.

Choose one between the two original equations: y=x+6 y = -x + 6

y=(23)+6y = -\left(\frac{2}{3}\right) + 6y=23+183y = -\frac{2}{3} + \frac{18}{3}y=163y = \frac{16}{3}

The intersection point of the lines y=x+6y = -x + 6 and y=2x+4y = 2x + 4 is (23,163).\left(\frac{2}{3}, \frac{16}{3}\right).

Dr Rahil Sachak-Patwa avatar
Written by: Dr Rahil Sachak-Patwa
LinkedIn
Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.

Hire a tutor

Please fill out the form and we'll find a tutor for you.

1/2 About yourself
Still have questions?
Let's get in touch.