Parallel and perpendicular lines
Two streets in a city can run in the same direction (they are parallel) or cross at a right angle (they are perpendicular). To find out without having to draw a map, all you need to do is look at one number for each line: its gradient.
The gradient of a line
Calculate the gradient of the line passing through points $A(2, -1)$ and $B(5, 8)$, and deduce from it what the gradient of any line perpendicular to it would be.
Calculate the gradient of the line passing through points $A(2, -1)$ and $B(5, 8)$, and deduce from it what the gradient of any line perpendicular to it would be.
Parallel lines
Given the lines $r: y = -2x + 5$ and $s: y = \frac{1}{2}x - 3$, state with reasons whether they are parallel, perpendicular, or intersecting but non-perpendicular.
Given the lines $r: y = -2x + 5$ and $s: y = \frac{1}{2}x - 3$, state with reasons whether they are parallel, perpendicular, or intersecting but non-perpendicular.
Find the explicit equation of the line that is parallel to $y = -3x + 2$ and passes through the point $P(0, 5)$.
Find the explicit equation of the line that is parallel to $y = -3x + 2$ and passes through the point $P(0, 5)$.
Perpendicular lines
Write down the gradient of a line that is parallel to and the gradient of a line that is perpendicular to the line with equation $y = 4x - 7$.
Write down the gradient of a line that is parallel to and the gradient of a line that is perpendicular to the line with equation $y = 4x - 7$.
Determine the point-slope equation of the line perpendicular to $y = \frac{2}{5}x - 1$ passing through the point $A(4, 2)$.
Determine the point-slope equation of the line perpendicular to $y = \frac{2}{5}x - 1$ passing through the point $A(4, 2)$.
Determine the value of parameter $m$ so that lines $r: mx + 4y - 1 = 0$ and $s: 2x - 5y + 8 = 0$ are perpendicular.
Determine the value of parameter $m$ so that lines $r: mx + 4y - 1 = 0$ and $s: 2x - 5y + 8 = 0$ are perpendicular.
Parallelism and perpendicularity in general form
Check analytically whether the lines given in general form $r: 3x - 6y + 2 = 0$ and $s: x - 2y - 5 = 0$ are parallel.
Check analytically whether the lines given in general form $r: 3x - 6y + 2 = 0$ and $s: x - 2y - 5 = 0$ are parallel.
Calculate the value of parameter $k$ so that the line $kx - 3y + 6 = 0$ is parallel to the line joining points $A(1, 2)$ and $B(4, 11)$.
Calculate the value of parameter $k$ so that the line $kx - 3y + 6 = 0$ is parallel to the line joining points $A(1, 2)$ and $B(4, 11)$.
The adjacent sides of a rectangular plot lie on the lines $r: 2x - 4y + 7 = 0$ and $s: 4x + 2y - 3 = 0$. Check analytically that the sides form a right angle. Then, find…
The adjacent sides of a rectangular plot lie on the lines $r: 2x - 4y + 7 = 0$ and $s: 4x + 2y - 3 = 0$. Check analytically that the sides form a right angle. Then, find the equation of a line parallel to $r$ that passes through the origin.
Applications: the perpendicular bisector
The perpendicular bisector of a segment $AB$ is the line perpendicular to $AB$ that passes through its midpoint. To find it, you combine two ideas you already know: the midpoint and the perpendicular gradient.
Find the explicit equation of the perpendicular bisector of the segment with endpoints $A(-2, 3)$ and $B(4, 1)$. (Remember that the perpendicular bisector is the line…
Find the explicit equation of the perpendicular bisector of the segment with endpoints $A(-2, 3)$ and $B(4, 1)$. (Remember that the perpendicular bisector is the line perpendicular to the segment that passes through its midpoint).
A ship sails in a straight line following the path $r: 3x + 4y - 24 = 0$. A lighthouse is located at point $F(7, 7)$. Calculate the shortest distance from the lighthouse…
A ship sails in a straight line following the path $r: 3x + 4y - 24 = 0$. A lighthouse is located at point $F(7, 7)$. Calculate the shortest distance from the lighthouse to the ship's path. (Hint: find the line perpendicular to $r$ that passes through $F$, find the point of intersection and calculate the distance between both points).