Further equations of lines
Point–gradient form, the line through two points, parallel and perpendicular lines through a point, and testing whether three lines are concurrent.
Point–gradient form
This is the quickest way to write a line when you know one point and the gradient.
Find the line through $(2, -3)$ with gradient $4$.
The line through two points
Find the line through $A(-1, 5)$ and $B(3, -3)$.
Drag A and B. The dashed line is perpendicular to AB and passes through A.
Make $AB$ horizontal. What does the perpendicular become?
Find the line through $(4, 1)$ perpendicular to $2x - 3y + 6 = 0$. Give it in general form.
Intersections and concurrent lines
Lines are concurrent if they all pass through one point. To test three lines, find where two of them meet (simultaneous equations), then check that point in the third.
Are $y = x + 1$, $y = -x + 5$ and $y = 3x - 3$ concurrent?
The line through $(1, -2)$ with gradient $-5$ is $y = -5x + b$. Find $b$.
$y + 2 = -5(x - 1)$, so $y = -5x + 5 - 2 = -5x + 3$.
The line through $(0, 2)$ and $(4, 10)$ is:
$m = \dfrac{8}{4} = 2$ and the $y$-intercept is $2$.
A line perpendicular to $y = -\tfrac13 x + 2$ through the origin is:
Negative reciprocal of $-\tfrac13$ is $3$.
Where do $y = 2x - 1$ and $y = -x + 5$ meet? Enter the $x$-coordinate.
$2x - 1 = -x + 5$, so $x = 2$.
The line through $(3, 4)$ parallel to $y = 5x$ is:
$y - 4 = 5(x - 3)$, so $y = 5x - 11$.
Lines are concurrent when they:
That is the definition. Lines with the same $y$-intercept are one example, but not the only one.
For which $k$ does $y = kx + 7$ pass through $(2, 3)$?
$3 = 2k + 7$, so $k = -2$.
The vertical line through the intersection of $y = x$ and $y = 6 - x$ is:
They meet at $(3, 3)$.
New numbers every time. Keep going until you get five in a row.
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