This page reviews the main ideas behind the slope of a line and the different forms of linear equations. After a concise summary of the theory, you will find carefully selected standard and advanced practice questions with complete, step-by-step solutions.
If a line passes through two distinct points \( P_1(x_1, y_1) \) and \( P_2(x_2, y_2) \), its slope is defined by: \[ m = \dfrac{y_2 - y_1}{x_2 - x_1}, \qquad x_2 \neq x_1 \]
The general (or standard) form of the equation of a straight line is: \[ Ax + By = C \] where \(A\), \(B\), and \(C\) are constants, and \(A\) and \(B\) are not both zero. Every straight line in the Cartesian plane can be represented in this form.
If the slope \(m\) of a line is known, its equation can be written as: \[ y = mx + b \] where \(m\) is the slope and \(b\) is the \(y\)-intercept.
An equation of a line with slope \(m\) passing through the point \(P(x_1, y_1)\) is: \[ y - y_1 = m(x - x_1) \]
If \(A = 0\) in the general equation \(Ax + By = C\), we obtain \(By = C \implies y = \dfrac{C}{B} = k\), which represents a horizontal line with slope \(0\).
If \(B = 0\), then \(Ax = C \implies x = \dfrac{C}{A} = h\), which represents a vertical line with undefined slope.
Two non-vertical lines are parallel if and only if their slopes are equal. Two non-vertical lines are perpendicular if and only if their slopes \(m_1\) and \(m_2\) satisfy: \[ m_1 m_2 = -1 \]
Find the slope of the line passing through the given points:
Find the equation of the line that passes through \((-2, 5)\) and has slope \(-4\).
Using the point–slope form:
\[ y - 5 = -4(x - (-2)) \implies y - 5 = -4(x + 2) \] \[ y - 5 = -4x - 8 \implies y = -4x - 3 \]Find the equation of the line passing through \((0, -1)\) and \((3, 5)\).
First, compute the slope:
\[ m = \dfrac{5 - (-1)}{3 - 0} = \dfrac{6}{3} = 2 \]Using the point–slope form with point \((0, -1)\):
\[ y - (-1) = 2(x - 0) \implies y + 1 = 2x \implies y = 2x - 1 \]Find the slope of the line given by:
\[ -2x + 4y = 6 \]Rewrite the equation in slope–intercept form (\(y = mx + b\)):
\[ 4y = 2x + 6 \implies y = \dfrac{2}{4}x + \dfrac{6}{4} \implies y = \dfrac{1}{2}x + \dfrac{3}{2} \]The slope is \( m = \dfrac{1}{2} \).
Find the equation of the line that passes through the intersection of the lines \(2x + y = 5\) and \(x - 3y = -1\), and is perpendicular to the line \(3x - 2y = 7\).
Step 1: Find the intersection point of the two lines.
From the second equation, express \(x\) in terms of \(y\):
\[ x = 3y - 1 \]Substitute this into the first equation (\(2x + y = 5\)):
\[ 2(3y - 1) + y = 5 \implies 6y - 2 + y = 5 \implies 7y = 7 \implies y = 1 \]Now find \(x\):
\[ x = 3(1) - 1 = 2 \]The intersection point is \( (2, 1) \).
Step 2: Find the slope of the perpendicular line.
Rewrite the given line \(3x - 2y = 7\) in slope-intercept form:
\[ 2y = 3x - 7 \implies y = \dfrac{3}{2}x - \dfrac{7}{2} \]Its slope is \(m_1 = \dfrac{3}{2}\). The slope of the line perpendicular to it is the negative reciprocal:
\[ m_2 = -\dfrac{1}{m_1} = -\dfrac{2}{3} \]Step 3: Write the equation of the required line.
Using the point-slope form with point \((2, 1)\) and slope \(m = -\dfrac{2}{3}\):
\[ y - 1 = -\dfrac{2}{3}(x - 2) \implies y - 1 = -\dfrac{2}{3}x + \dfrac{4}{3} \implies y = -\dfrac{2}{3}x + \dfrac{7}{3} \]In general form: \(2x + 3y = 7\).
Find the equation of the perpendicular bisector of the line segment joining the points \(A(-1, 3)\) and \(B(5, 7)\).
Step 1: Find the midpoint of segment \(AB\).
\[ M = \left(\dfrac{-1 + 5}{2}, \dfrac{3 + 7}{2}\right) = \left(\dfrac{4}{2}, \dfrac{10}{2}\right) = (2, 5) \]Step 2: Find the slope of line segment \(AB\).
\[ m_{AB} = \dfrac{7 - 3}{5 - (-1)} = \dfrac{4}{6} = \dfrac{2}{3} \]Step 3: Find the slope of the perpendicular bisector.
\[ m_{\perp} = -\dfrac{1}{m_{AB}} = -\dfrac{3}{2} \]Step 4: Write the equation using point-slope form.
Using point \((2, 5)\) and slope \(m = -\dfrac{3}{2}\):
\[ y - 5 = -\dfrac{3}{2}(x - 2) \implies y - 5 = -\dfrac{3}{2}x + 3 \implies y = -\dfrac{3}{2}x + 8 \]In standard form: \(3x + 2y = 16\).
Find the coordinates of the reflection (image) of the point \(P(2, -3)\) across the line \(L: x + 2y = 4\).
Step 1: Find the slope of line \(L\).
Rewrite \(x + 2y = 4\) in slope-intercept form:
\[ 2y = -x + 4 \implies y = -\dfrac{1}{2}x + 2 \]The slope of \(L\) is \(m_L = -\dfrac{1}{2}\).
Step 2: Find the equation of the line passing through \(P(2, -3)\) perpendicular to \(L\).
The perpendicular slope is \(m_{\perp} = -\dfrac{1}{-1/2} = 2\). Using point \(P(2, -3)\):
\[ y - (-3) = 2(x - 2) \implies y + 3 = 2x - 4 \implies y = 2x - 7 \]Step 3: Find the intersection point (foot of the perpendicular, \(Q\)) between \(L\) and the perpendicular line.
Substitute \(y = 2x - 7\) into \(x + 2y = 4\):
\[ x + 2(2x - 7) = 4 \implies x + 4x - 14 = 4 \implies 5x = 18 \implies x = \dfrac{18}{5} = 3.6 \]Now solve for \(y\):
\[ y = 2\left(\dfrac{18}{5}\right) - 7 = \dfrac{36}{5} - \dfrac{35}{5} = \dfrac{1}{5} = 0.2 \]The foot of the perpendicular is \(Q\left(\dfrac{18}{5}, \dfrac{1}{5}\right)\).
Step 4: Find the reflected point \(P'(x', y')\).
Since \(Q\) is the midpoint between \(P(2, -3)\) and \(P'(x', y')\):
\[ \dfrac{2 + x'}{2} = \dfrac{18}{5} \implies 2 + x' = \dfrac{36}{5} \implies x' = \dfrac{36}{5} - \dfrac{10}{5} = \dfrac{26}{5} = 5.2 \] \[ \dfrac{-3 + y'}{2} = \dfrac{1}{5} \implies -3 + y' = \dfrac{2}{5} \implies y' = \dfrac{2}{5} + \dfrac{15}{5} = \dfrac{17}{5} = 3.4 \]The coordinates of the reflected point are \(P'\left(\dfrac{26}{5}, \dfrac{17}{5}\right)\) or \((5.2, 3.4)\).