Parallel and perpendicular lines, intersection of 2 straight lines, solving a two-variable system of equations
The lines L1: y=x+1 and L2: y=−2x+7 are drawn on the grid below. They cross at one point, (2,3).
Every point on L1 satisfies the equation of L1, and every point on L2 satisfies the equation of L2. The point (2,3) lies on both lines, so it satisfies both equations at once. Putting x=2 and y=3 into each:
Substituting x=2 into either equation gives y=3, so the same pair of values makes both equations true. No other pair does: any other point of L1 is off L2, so it fails the second equation, and any other point of L2 fails the first.
Two equations that must hold at the same time form a system of equations. A solution of the system is a pair of values of x and y that satisfies both equations.
When each equation is the equation of a line, a solution is a point lying on both lines: the point where they intersect.
A system of equations can have any number of equations and any number of unknowns. When the equations describe straight lines, the system has two equations and two unknowns, x and y, and is called a 2×2 system.
Written as a system, the two lines above are
and the solution is x=2, y=3.
The diagram shows three lines and their intersections P, Q and R.
Which point is the solution of the system
The solution is the point where the two lines of the system cross, so the first step is to find which two of the drawn lines the equations describe. Rearranging each into the form y=mx+c shows its gradient and y-intercept:
The first line rises 3 for every 2 across and crosses the y-axis at 1.5: that is L2. The second falls 2 for every 1 across and crosses the y-axis at 5: that is L3. The system's lines are L2 and L3, which cross at Q.
Reading the grid, Q is (1,3). Checking in both equations: 3(1)−2(3)+3=0 and 2(1)+3−5=0. So the solution is x=1, y=3, the point Q.
Checkpoint 1
The following diagram shows three straight lines and their intersections P, Q and R.
Which point represents the solution to the following system of equations?
A graphic display calculator solves a system of linear equations directly. Enter the two equations, or their coefficients, into its system solver and it returns x and y. That is the route to take when the coefficients do not cancel neatly and the answer is wanted to a given accuracy.
Checkpoint 2
The lines 3x+7y−5=0 and 5x+2y−20=0 intersect at the point Q.
Use technology to find the coordinates of Q, giving each coordinate correct to three significant figures.
Two lines in the same plane are parallel when they never meet, however far they are extended in either direction.
The two lines above have equations y=2x−4 and y=2x+1. On each of them, moving 1 to the right moves 2 up, so both have gradient 2. Two lines with the same gradient climb at the same rate, so the vertical gap between them stays the same all the way along and never closes: they are parallel. The reverse holds too. If one line climbed faster than the other, the gap between them would shrink and they would eventually meet, so parallel lines must have the same gradient.
Two distinct lines with gradients m1 and m2 are parallel exactly when
Two lines with the same gradient are either parallel or the same line, and their y-intercepts decide which: distinct parallel lines have the same gradient and different y-intercepts.
Vertical lines have no gradient. They are all parallel to one another, and a vertical line crosses every non-vertical line exactly once.
Find the equation of the line through (2,−1) that is parallel to y=−3x+2. Give the answer in the form y=mx+c.
The given line has gradient −3, so the required line has gradient −3 as well, and its equation is y=−3x+c for some value of c. The point (2,−1) lies on it, so substituting x=2 and y=−1 finds c:
The line is y=−3x+5. It has the same gradient as the given line and a different intercept, 5 against 2, so the two are distinct parallel lines.
Checkpoint 3
The line L is parallel to y=4x−3, and passes through (4,3).
Find the equation of L in the form y=mx+c.
Solving the equations of two parallel lines as a system shows what happens when there is no intersection to find. Equating y=2x−4 and y=2x+1 gives
which is false whatever x is. No pair of values satisfies both equations: the system has no solution, because the lines have no point in common.
Two equations can also describe one and the same line. The equation x−2y−2=0 rearranges to y=21x−1, so as a system with y=21x−1 it says the same thing twice. Substituting gives
which is true whatever x is. Every point of the line is a solution, so the system has infinitely many solutions. Two equations that describe the same line are called coincident.
The diagram above illustrates the three possible relationships between two straight lines, and each one fixes how many solutions the system of their equations has.
Gradients and intercepts can only be compared when both equations are in the same form, so an equation given in general form or point-gradient form is rearranged to y=mx+c before comparing.
Checkpoint 4
The line L has equation y=2x−3. Consider the following lines:
Classify each line by its relation to L.
Checkpoint 5
Consider the system of equations
Complete the sentence.
Two lines are perpendicular when they meet at a right angle.
Lines ① and ② above cross at the marked point at a right angle. Follow line ① for 4 units in the positive x-direction and it rises 3 units in the positive y-direction, so its gradient is 43. Line ②, which is perpendicular to ①, has the rise and the run the other way round: for 3 units in the positive x-direction it falls 4 units, a change of −4 in y, so its gradient is
Two lines that meet at a right angle are a quarter turn apart, so a quarter turn about their crossing point carries one onto the other. Draw the right triangle whose longest side lies along the first line, with a horizontal side of 4 and a vertical side of 3, and turn the whole triangle a quarter turn clockwise about the crossing.
Its longest side lands on the second line. The side that was horizontal, of length 4, now stands vertical and points downwards, and the side that was vertical, of length 3, now runs across. The run and the rise have traded places, and the one that now points down has changed sign: a gradient of 43 has become −34.
The same turn works for any line: a line of gradient ba is perpendicular to a line of gradient −ab, and the product of the two gradients is
Two lines with gradients m1 and m2 are perpendicular exactly when
so the gradient of a line perpendicular to one with gradient m is −m1: the reciprocal of m with its sign changed.
A horizontal line has gradient 0 and a vertical line has no gradient, so their product cannot be formed. They are perpendicular all the same: every horizontal line is perpendicular to every vertical line, and to no other line. That pair is the one case the rule does not cover.
Checkpoint 6
Four lines are given by their equations:
Match each line with the lines that are perpendicular to it.
Checkpoint 7
Consider the three lines drawn on the grid below.
Select every statement that is true.
Find the equation of the line through (4,−3) that is perpendicular to y=52x+2. Give the answer in the form y=mx+c.
The given line has gradient 52. The required gradient m satisfies 52m=−1, so m=−25 and the line is y=−25x+c for some value of c. The point (4,−3) lies on it, so substituting x=4 and y=−3 finds c:
The line is y=−25x+7, and 52×(−25)=−1, so the two lines are perpendicular.
Checkpoint 8
The line L is perpendicular to the line y=21x−3, and intersects it at x=1.
Find the equation of L in the form y=mx+c.
Nice work reaching the end of Line intersections, parallel lines, and perpendicular lines. Here is what this lesson set out to teach - worth reading back with fresh eyes:
0/8 checkpoints checked off