Coordinate geometry flashcards

Test yourself one card at a time: say the answer, flip, then mark it. Cards you know drop out; the rest come back until you know them. Your progress stays in this browser.

How do you find the gradient of the line through \((x_1,y_1)\) and \((x_2,y_2)\)?
\(m=\frac{y_2-y_1}{x_2-x_1}\), the change in \(y\) divided by the change in \(x\).
Find the gradient of the line through \((1,2)\) and \((5,10)\).
\(m=\frac{10-2}{5-1}=\frac84=2\).
What does a negative gradient tell you about a line?
The line slopes downwards from left to right: as \(x\) increases, \(y\) decreases. A horizontal line has gradient 0; a vertical line has undefined gradient.
How do you find the midpoint of a line segment?
Average the coordinates: midpoint \(=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\).
Find the midpoint of \(A(-2,7)\) and \(B(6,1)\).
\(\left(\frac{-2+6}{2},\frac{7+1}{2}\right)=(2,4)\).
How do you find the length of a line segment between two points?
Use Pythagoras with the horizontal and vertical differences: \(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\).
Find the distance between \((1,2)\) and \((7,10)\).
\(d=\sqrt{6^2+8^2}=\sqrt{100}=10\).
State the equation of a straight line in gradient-intercept form and say what each letter means.
\(y=mx+c\), where \(m\) is the gradient and \(c\) is the \(y\)-intercept, the value of \(y\) when \(x=0\).
How do you find the equation of a line given its gradient and one point on it?
Substitute \(m\) and the point \((x,y)\) into \(y=mx+c\) and solve for \(c\). Or use \(y-y_1=m(x-x_1)\).
Find the equation of the line through \((0,3)\) and \((2,9)\).
\(m=\frac{9-3}{2-0}=3\) and the \(y\)-intercept is 3, so \(y=3x+3\).
What can you say about the gradients of two parallel lines?
They are equal. E.g. \(y=2x+1\) and \(y=2x-5\) are parallel because both have gradient 2.
(Extended) What is the relationship between the gradients of two perpendicular lines?
Their product is \(-1\): if one has gradient \(m\), the other has gradient \(-\frac1m\). E.g. 2 and \(-\frac12\).
(Extended) Find the equation of the line perpendicular to \(y=2x+1\) passing through \((4,3)\).
Perpendicular gradient \(=-\frac12\). Then \(3=-\frac12(4)+c\), so \(c=5\). The line is \(y=-\frac12x+5\).
Write \(2x+3y=12\) in the form \(y=mx+c\) and state the gradient.
\(3y=-2x+12\), so \(y=-\frac23x+4\). The gradient is \(-\frac23\) and the \(y\)-intercept is 4.
How do you find where a line \(ax+by=c\) crosses the axes?
Put \(x=0\) to get the \(y\)-intercept, and put \(y=0\) to get the \(x\)-intercept. E.g. \(2x+3y=12\) crosses at \((0,4)\) and \((6,0)\).
What is the equation of a horizontal line and of a vertical line?
Horizontal: \(y=k\) (gradient 0). Vertical: \(x=k\) (gradient undefined). Both are constants through the point \(k\) on the relevant axis.
How do you find the point where two lines intersect?
Solve their equations simultaneously, by substitution or elimination. The solution \((x,y)\) is the intersection point, which is also where the graphs cross.
Do the points \((1,3)\), \((2,5)\) and \((4,9)\) lie on one straight line?
Gradient of first two: \(\frac{5-3}{2-1}=2\). Gradient of last two: \(\frac{9-5}{4-2}=2\). The gradients match, so they are collinear.
On a distance-time graph, what does the gradient represent, and what does a horizontal line mean?
The gradient is the speed (\(\frac{\text{distance}}{\text{time}}\)). A horizontal line means the object is stationary.
(Extended) On a speed-time graph, what do the gradient and the area under the graph represent?
The gradient is the acceleration (a negative gradient is deceleration). The area under the graph is the distance travelled.
Line \(L\) has equation \(y=-3x+4\). What is the gradient of a line parallel to \(L\)?
\(-3\), because parallel lines have the same gradient.
\(M\) is the midpoint of \(AB\), where \(A=(1,2)\) and \(M=(4,6)\). Find \(B\).
Double the step from \(A\) to \(M\): \(x\) goes up 3, so \(x_B=7\); \(y\) goes up 4, so \(y_B=10\). \(B=(7,10)\).
How do you decide if a point lies on a given line?
Substitute its coordinates into the equation. If both sides are equal, the point is on the line. E.g. \((3,7)\) is on \(y=2x+1\) as \(7=2(3)+1\).
Find the gradient of the line through \((-3,4)\) and \((1,-4)\).
\(m=\frac{-4-4}{1-(-3)}=\frac{-8}{4}=-2\).
Find the length of the line joining \((-2,1)\) and \((1,5)\).
\(d=\sqrt{3^2+4^2}=\sqrt{25}=5\).
(Extended) Find the perpendicular bisector of the segment joining \((0,0)\) and \((4,2)\).
Midpoint is \((2,1)\). Segment gradient is \(\frac12\), so perpendicular gradient is \(-2\). Then \(1=-2(2)+c\), so \(c=5\): \(y=-2x+5\).
What is the \(y\)-intercept of \(y=4x-7\), and where does it cross the \(x\)-axis?
The \(y\)-intercept is \(-7\). Put \(y=0\): \(4x=7\), so \(x=\frac74=1.75\).

Practise this topicNotes Questions (179)Test builder