You will probably recognise the most commonly used form of the equation of a straight line: y = mx + c where m represents the gradient and c represents the y-intercept. It is important not to overlook this topic, as it comes up a lot in work on Functions, Vectors and Calculus. Make sure that you can use all three forms of the equation of a straight line, as this will help you with the Equation of Tangent and Normal.
On this page, you should learn to
- Use the three different forms of the equations of straight lines
- \(y = mx + c \)
- \(ax+by + d = 0 \)
- \(y-y_1=m(x-x_1) \)
- Solve problems with parallel lines \(m_1=m_2\)
- Solve problems with perpendicular lines \(m_1 \times m_2=-1\)
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Here is a quiz that practises the skills from this page
START QUIZ!
What is the gradient and y intercept of the following line?
gradient =
y intercept =
gradient = \(\frac{rise}{run}\)
The equation of the following line can be written in the form \(y=ax+b\)
Find a and b
a =
b =
a = gradient
b = y intercept
Which of the following is the correct equation of the following straight line
gradient = 3
y intercept = -1
Which of the following lines passes through the point (2 , 1)
y = -x + 3
1 = -2 + 3
Which of the following lines is parallel to \(y = \frac{1}{2}x+2\)
x - 2y + 3 = 0 can be rearranged into the form y = mx + c
x + 3 = 2y
\(y = \frac{1}{2}x+\frac{3}{2}\)
Which of he following gradients could represent the gradients of 2 perpendicular lines
A) 2
B) \(\frac{2}{5}\)
C) 5
D) \(-\frac{5}{2}\)
If a line has gradient m then a perpendicular line has gradient \(\frac{-1}{m}\)
What is the gradient of the line that is perpendicular to the line below
gradient =
The gradient a line perpendicular to \(y = -\frac{2}{3}x-2\) can be written in the form \(\frac{a}{b}\)
Find a and b
a =
b =
If a line has gradient m then a perpendicular line has gradient \(\frac{-1}{m}\)
The equation line below can be written in the form \(ax+by+c=0\)
Find a , b and c
a =
b =
c =
gradient = \(-\frac{2}{5}\)
y intercept = -2
\(y = -\frac{2}{5}x-2\\ 5y=-2x-10\\ 5y+2x+10=0\\ 2x+5y+10=0\)
The equation of the line that is perpendicular to the line below and that passes through the point (1,1) can be written in the form \(ax+2y+b=0\)
Find a and b
a =
b =
gradient of line in graph = \(-\frac{2}{5}\)
gradient of perpendicular line = \(\frac{5}{2}\)
\(y = \frac{5}{2}x+c \)
passes through (1,1)
\(1 = \frac{5}{2}\cdot1+c\\ -\frac{3}{2} = c \\ y = \frac{5}{2}x-\frac{3}{2} \\ 2y=5x-3\\ -5x+2y+3=0 \)
Here is a quiz about parallel and perpendicular lines
START QUIZ!
Match up the gradients with the equations of the straight lines
-2 -0.5 0.5 2
y = 2x + 1 , gradient =
2x + y + 1 = 0 , gradient =
\(y-1= \frac{1}{2}(x+2)\) , gradient =
x + 2y - 1 = 0 , gradient =
y = mx + c , gradient = m
y = 2x + 1 , gradient = 2
2x + y + 1 = 0
y = -2x - 1 , gradient = -2
\(y-1= \frac{1}{2}(x+2)\) , gradient = 0.5
x + 2y - 1 = 0
y = -0.5x + 0.5 , gradient = -0.5
The equation of a line that is parallel to y = 3x - 2 and passes through the point (1 , 2) is given by
y = mx + c
Find m and c
m =
c =
If the line is parallel to y = 3x - 2 , then the gradient = 3
We can write it in the form
y = 3x + c
We know that it passes through (1 , 2)
2 = 3(1) + c
c = -1
Therefore, the equation is y = 3x - 1
The equation of a line that is parallel to \(y=\frac{2}{3}x+1\) and passes through (3 , -1) is given by
ax + by + d = 0 , where a > 0
Find a , b and d
a =
b =
d =
If the line is parallel to \(y=\frac{2}{3}x+1\) , then the gradient = \(\frac{2}{3}\)
We can write it in the form
\(y=\frac{2}{3}x+c\)
We know that it passes through (3 , -1)
-1 = \(\frac{2}{3}\)(3) + c
c = -3
The equation is y = \(\frac{2}{3}\)x - 3
Multiply through by 3 and rearrange to put in the correct form:
3y = 2x - 9
2x - 3y - 9 = 0
The equation of the line \(l_1\) is given by 2x + 3y - 5 = 0
\(l_2\) is parallel to \(l_1\)
Which of the following could be the equation of the line \(l_2\)
2x + 3y - 5 = 0
\(y = -\frac{2}{3}x+\frac{5}{3}\)
Therefore the gradients of the lines \(l_1\) and \(l_2\) are \(-\frac{2}{3}\)
Rearrange 4x + 6y - 1 = 0 to find the gradient
\(y = -\frac{4}{6}x+\frac{1}{6}\)
The gradient of this line is \(-\frac{4}{6}=-\frac{2}{3}\)
A straight line, \(l\) has gradient = \(\frac{1}{2}\)
What is the gradient of a line perpendicular to \(l\)?
gradient =
If the gradients of the lines are \(m_1\) and \(m_2\), then \(m_1 \times m_2=-1\)
\(\frac{1}{2} \times m_2=-1\)
\(\frac{1}{2} \times (-2)=-1\)
The gradient of \(l_1\) is \(m_1\)
The gradient of \(l_2\) is \(m_2\)
What is \(m_1 \times m_2\)
\(m_1 \times m_2\) =
Since these lines are perpendicular, then \(m_1 \times m_2=-1\)
What is the gradient of a line that is perpendicular to the one in the graph?
gradient =
The gradient of the line in the graph is \(-\frac{2}{3}\)
The gradient of the perpendicular line = \(\frac{-1}{-\frac{2}{3}}=\frac{3}{2}\)
the equation of the line \(l_1\) is given by
y - 3 = -2 (x - 2)
\(l_2\) is perpendicular to \(l_1\)
What is the gradient of \(l_2\)?
gradient =
gradient of \(l_1\)= -2
gradient of \(l_2=-\frac{1}{-2}=\frac{1}{2}\)
The equation of the line \(l_1\) is given by 2x + 5y - 5 = 0
\(l_2\) is perpendicuar to \(l_1\)
Which of the following could be the equation of the line \(l_2\)
2x + 5y - 5 = 0
\(y = -\frac{2}{5}x+1\)
The gradient of the line \(l_1\)= \(-\frac{2}{5}\)
Therefore, the gradient of the line \(l_2\)= \(\frac{5}{2}\)
Rearrange 5x - 2y + 3 = 0 to find the gradient
2y = 5x + 3
\(y=\frac{5}{2}x+\frac{3}{2}\)
The gradient of this line is \(\frac{5}{2}\)
The lines \(l_1,l_2, l_3 \) and \( l_4\) form a square.
\( l_1:\) x + 2y = 12
\( l_2:\) 2x - y = -6
\( l_3: \ y=-\frac{1}{2}x+\frac{7}{2}\)
Which of the following lines completes the square?
gradient of \( l_1\)= \(-\frac{1}{2}\)
gradient of \( l_2\)= 2
gradient of \( l_3\)= \(-\frac{1}{2}\)
Therefore, gradient of \( l_4\)= 2
There are two possible solutions:
The line \(l_1\) has equation 2x+5y+6=0
The line \(l_2\) is perpendicular to the line \(l_1\)and passes through the point (2 , -2)
a) Find the equation of \(l_2\)in the form ax+by+d=0, where a, b and d are constants.
b) Find the coordinates where \(l_2\) meets the y axis
Hint
Full Solution
The point A has coordinates (a , 3) and the point B has coordinates (7 , b).
The line AB has equation 2x + 3y = 11.
a) Find the values of a and b
The line AC is perpendicular to the line AB.
b) Find the equation of the line AC in the form ax + by + d = 0, where a, b and d are constants
c) Given that C lies on the x axis, find its coordinates.
Hint
Full Solution
The line \(l_1 \) passes through the point P(3k , 2k) with gradient = -2.
\(l_1 \) meets the x axis at A and the y axis at B.
a) Find the equation of the line \(l_1 \) and show that A(4k , 0)
b) Find the area of the triangle AOB in terms of k
The line \(l_2\) passes through P and is perpendicular to \(l_1 \)
c) Find the equation of \(l_2\)
\(l_2\)meets the x axis at C
d) Show that the midpoint of PC lies on the line y = x
Hint
Full Solution
How much of Equation of a Straight Line have you understood?
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