Parallel and perpendicular lines
The streets of a city laid out on a grid are either parallel or perpendicular to each other: they never cross at an angle for no reason. Being able to recognise and construct parallel or perpendicular lines from their gradient or direction vector is the piece that completes everything you have learnt in coordinate geometry.
Parallel lines
Given line $r$ with equation $y = -5x + 3$, state the gradient of any line parallel to $r$ and the gradient of any line perpendicular to $r$.
Given line $r$ with equation $y = -5x + 3$, state the gradient of any line parallel to $r$ and the gradient of any line perpendicular to $r$.
Determine the relative position of lines $r: 2x - y + 4 = 0$ and $s: -4x + 2y - 8 = 0$.
Determine the relative position of lines $r: 2x - y + 4 = 0$ and $s: -4x + 2y - 8 = 0$.
Perpendicular lines
Write the coordinates of a vector that is parallel to, and another that is perpendicular to, vector $\vec{u} = (-4, 7)$.
Write the coordinates of a vector that is parallel to, and another that is perpendicular to, vector $\vec{u} = (-4, 7)$.
Determine the general equation of the line perpendicular to $3x + 4y - 2 = 0$ that passes through the origin.
Determine the general equation of the line perpendicular to $3x + 4y - 2 = 0$ that passes through the origin.
Two lines have gradients $m_1 = \frac{2}{3}$ and $m_2 = \frac{3}{2}$. A student concludes that, since they are formed by inverted digits, the lines are perpendicular.…
Two lines have gradients $m_1 = \frac{2}{3}$ and $m_2 = \frac{3}{2}$. A student concludes that, since they are formed by inverted digits, the lines are perpendicular. Explain the mistake in this "trick" and what the value of $m_2$ should be for the lines to be truly perpendicular.
The perpendicular bisector of a line segment
Calculate the general equation of the perpendicular bisector of the line segment whose endpoints are points $A(-3, 5)$ and $B(1, -1)$.
Calculate the general equation of the perpendicular bisector of the line segment whose endpoints are points $A(-3, 5)$ and $B(1, -1)$.
Applications: altitudes, reflections and shapes
Parallel and perpendicular lines solve many geometry problems: the altitude of a triangle from a vertex is perpendicular to the opposite side; the reflection of $A$ in a line $r$ is found by drawing the perpendicular to $r$ through $A$ and reflecting to the same distance on the other side; and checking that the sides of a quadrilateral are parallel (or perpendicular) in pairs lets you prove that it is a parallelogram, a rectangle or a rhombus.
In the triangle with vertices $A(-2, 1)$, $B(4, 3)$ and $C(2, -3)$, find the general equation of the altitude from vertex $A$.
In the triangle with vertices $A(-2, 1)$, $B(4, 3)$ and $C(2, -3)$, find the general equation of the altitude from vertex $A$.
Find the coordinates of the orthocentre (intersection point of the altitudes) of the triangle whose vertices are $A(0, 4)$, $B(-4, 0)$ and $C(2, -2)$.
Find the coordinates of the orthocentre (intersection point of the altitudes) of the triangle whose vertices are $A(0, 4)$, $B(-4, 0)$ and $C(2, -2)$.
Prove analytically that the quadrilateral with vertices $A(-1, -2)$, $B(4, -1)$, $C(5, 4)$ and $D(0, 3)$ is a rhombus. To do this, check that opposite sides are parallel…
Prove analytically that the quadrilateral with vertices $A(-1, -2)$, $B(4, -1)$, $C(5, 4)$ and $D(0, 3)$ is a rhombus. To do this, check that opposite sides are parallel and that its diagonals are perpendicular to each other.