Coordinate Geometry

The first type is pure reading of the plane. Three kinds of questions come here: whether a point lies on a given line, lines parallel to the axes, and the reflection (mirror image) of a point or a line.

1. Does the point lie on the line? As we saw in the foundation, put the coordinates in the equation. Nothing else needs to be drawn or calculated. Take y = 3x − 2. Is (2, 4) on it? 3 × 2 − 2 = 4, so yes. Is (1, 2) on it? 3 × 1 − 2 = 1 ≠ 2, so no. When one option says all of the above, check every point, because quite often all of them are correct.

[[fig:0]]

2. Lines parallel to the axes. If every point of a line has the same y-value, say y = 3, the line is horizontal: parallel to the x-axis and 3 units above it. If every point has the same x-value, say x = −2, the line is vertical: parallel to the y-axis and 2 units to its left.

• The x-axis itself is the line y = 0, and the y-axis itself is the line x = 0.
• A horizontal line y = b (b ≠ 0) never meets the x-axis. A vertical line x = a (a ≠ 0) never meets the y-axis.
• The lines x = a and y = b always meet at the point (a, b).

[[fig:1]]

3. Reflection (mirror image). Imagine a mirror lying along a line. The image is as far behind the mirror as the point is in front of it.

[[fig:2]]

An easy way to hold the table in your head: in the x-axis mirror the x-coordinate stays and y changes sign; in the y-axis mirror the y-coordinate stays and x changes sign. In the mirror y = x the two coordinates just swap places.

[[fig:3]]

Reflecting a whole line. If the mirror is the x-axis, every point (x, y) of the image comes from the point (x, −y) of the original line. So replace y by −y in the equation. For example, reflect y = 3x − 4 in the x-axis: −y = 3x − 4, that is, y = −3x + 4. In the y-axis mirror, replace x by −x instead, which gives y = −3x − 4.

[[fig:4]]

Distance between two points. Take A(x₁, y₁) and B(x₂, y₂). Draw a horizontal line from A and a vertical line from B. They meet at a corner and make a right triangle whose horizontal side is x₂ − x₁, whose vertical side is y₂ − y₁, and whose hypotenuse is AB. So by Pythagoras,

Formula (Distance formula):
>> AB = √(x₂ − x₁)² + (y₂ − y₁)²

That is the whole distance formula: Pythagoras on graph paper. The order of subtraction does not matter, because each difference is squared.

Radhika's home is at (2, −1) and her coaching centre at (8, 7). The horizontal gap is 6 and the vertical gap is 8, so the distance is √36 + 64 = 10 units. See the triplet 6–8–10 hiding there. Always look for Pythagorean triplets (3–4–5, 5–12–13, 8–15–17, 7–24–25 and their multiples); then you don't need the square root at all.

[[fig:0]]

The exam uses the distance formula in two more ways:

• Is a triangle isosceles, equilateral or right-angled? Find all three sides. Two equal sides mean isosceles; a² + b² = c² means right-angled.
• A point equidistant from two points: if P(x, 0) is equally far from A and B, write PA² = PB² and solve. The x² terms cancel and a simple linear equation remains.

Midpoint. The midpoint of AB is just the average of the coordinates:

Formula (Midpoint formula):
>> M = (x₁ + x₂2, y₁ + y₂2)

The midpoint of (−3, 4) and (9, 10) is (62, 142) = (3, 7). It also works backwards: if M and A are known, the other end is B = (2x_M − x₁, 2y_M − y₁).

[[fig:1]]

Section formula. If P divides AB internally in the ratio m : n (AP : PB = m : n), then P is a weighted average in which each end is multiplied by the part of the ratio that is away from it:

Formula (Section formula (internal division)):
>> P = (mx₂ + nx₁m + n, my₂ + ny₁m + n)

For example, let P divide A(1, 2) and B(7, 11) in the ratio 1 : 2. Then P = (1 × 7 + 2 × 13, 1 × 11 + 2 × 23) = (3, 5). A quick check: from A to B, x grows by 6, and P is 13 of the way along, so x = 1 + 2 = 3. It fits.

[[fig:2]]

For external division (P outside AB, AP : PB = m : n) the plus sign becomes minus: P = (mx₂ − nx₁m − n, my₂ − ny₁m − n).

In what ratio does an axis divide AB? The x-axis meets AB at a point with y = 0, so the ratio is −y₁ : y₂. In the same way the y-axis divides it in the ratio −x₁ : x₂. For A(2, −3) and B(5, 6), the x-axis divides AB in the ratio 3 : 6 = 1 : 2.

[[fig:3]]

Naming a figure with the distance formula. Some questions give three or four points and ask what figure they make. Work out the squared lengths first (no square roots are needed until the end) and read them:

• Three points: if the two smaller distances add up to the largest one, the points are collinear. Otherwise check for two equal sides (isosceles), three equal sides (equilateral), or a² + b² = c² (right-angled).
• Four points in order: all four sides equal with equal diagonals is a square; equal sides with unequal diagonals is a rhombus; opposite sides equal with equal diagonals is a rectangle; only opposite sides equal is a parallelogram (its diagonals have the same midpoint).

Radhika tests (0, 0), (3, 1), (4, 4) and (1, 3). Each side squared is 10, and the diagonals squared are 32 and 8, so she has a rhombus that is not a square.

A median of a triangle joins a vertex to the midpoint of the opposite side. Every triangle has three medians, and they always pass through one common point, the centroid G. Physically it is the balancing point of a triangular plate. That is why questions talk about placing a fountain, a scarecrow or a load "at the centroid".

Its formula is the simplest in the chapter. For A(x₁, y₁), B(x₂, y₂), C(x₃, y₃):

Formula (Centroid):
>> G = (x₁ + x₂ + x₃3, y₁ + y₂ + y₃3)

Add the three x's and divide by 3; add the three y's and divide by 3. It is an average, just as the midpoint was the average of two points.

Radhika's three friends live at (2, 7), (0, 1) and (7, 1). The point that balances all three homes is G = (2 + 0 + 73, 7 + 1 + 13) = (3, 3).

[[fig:0]]

The 2 : 1 property. The centroid divides every median in the ratio 2 : 1 from the vertex. So if AD is a median, AG = 23 AD and GD = 13 AD. In coordinates, G = A + 2D3 (apply it to x and y separately). Two quick reverse results follow:

• If the median AD is 12 cm, then AG = 8 cm and GD = 4 cm.
• If A and G are known, the midpoint D of BC is D = (3x_G − x_A2, 3y_G − y_A2).

[[fig:1]]

Third vertex from the centroid. This reverse question comes up again and again. Since x₁ + x₂ + x₃ = 3x_G, the missing vertex is
x₃ = 3x_G − x₁ − x₂, y₃ = 3y_G − y₁ − y₂
Say G = (2, 1) and two vertices are (5, −2) and (−3, 4). The third vertex = (6 − 5 + 3, 3 + 2 − 4) = (4, 1).

[[fig:2]]

Two more facts are worth knowing. The centroid always lies inside the triangle, and the triangle formed by joining the midpoints of the sides has the same centroid as the original triangle.

Apart from the centroid, a triangle has three more well-known centres. You already know what they mean from Geometry; here we only need their coordinates.

Circumcentre O. This is the point where the perpendicular bisectors of the sides meet. It is equidistant from all three vertices, so it is the centre of the circle passing through A, B and C. Whenever a question says "a point at equal distance from the three corners", it is asking for the circumcentre.

[[fig:0]]

Orthocentre H. This is the point where the three altitudes meet (the perpendiculars dropped from the vertices to the opposite sides).

Incentre I. This is where the angle bisectors meet. It is equidistant from the three sides, and that distance is the inradius r. In other words, the perpendicular distance from the incentre to any side is the inradius. Its coordinates are a weighted average with the side lengths as weights:

Formula (Incentre):
>> I = (ax₁ + bx₂ + cx₃a + b + c, ay₁ + by₂ + cy₃a + b + c), where a = BC, b = CA, c = AB (each side is the one opposite its vertex).

[[fig:1]]

Right-angled triangles make this type easy. In a right-angled triangle:

• the circumcentre is the midpoint of the hypotenuse, and the orthocentre is the right-angled vertex;
• the reason: the hypotenuse is a diameter of the circumcircle, and the two legs are themselves altitudes.

Radhika plots A(0, 0), B(9, 0), C(0, 5). The angle at A is 90°, because AB lies on the x-axis and AC on the y-axis. So the orthocentre is A(0, 0), and the circumcentre is the midpoint of BC, which is (4.5, 2.5).

[[fig:2]]

How do you know a triangle is right-angled? If two sides lie along the axes, the right angle is at the origin. Otherwise look at the slopes: two sides whose slopes multiply to −1 meet at a right angle (slope comes in Type 6). You can also check a² + b² = c² with the distance formula.

Equilateral triangle: all four centres (G, O, H, I) coincide.

The Euler line. In any triangle H, G and O lie on one straight line, and G sits between them with HG : GO = 2 : 1. So:

• G = H + 2O3 (when H and O are known)
• O = 3G − H2 (when G and H are known)
• H = 3G − 2O (when G and O are known)

Take H = (1, 7) and O = (4, 1). Then G = (1 + 83, 7 + 23) = (3, 3). The incentre is generally not on this line.

[[fig:3]]

In Mensuration you needed a base and a height for the area of a triangle. In coordinates you need neither; the three vertices are enough. For A(x₁, y₁), B(x₂, y₂), C(x₃, y₃):

Formula (Area of a triangle):
>> Area = 12 |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|

Read the pattern once and it stays with you: each x is multiplied by the difference of the other two y's, taken in cyclic order (2–3, 3–1, 1–2). The modulus sign is there because an area can never be negative. If the bracket comes out negative, drop the minus sign.

Radhika's triangle has vertices A(1, 1), B(5, 2), C(3, 6).
Area = 12 |1(2 − 6) + 5(6 − 1) + 3(1 − 2)| = 12 |−4 + 25 − 3| = 12 × 18 = 9 square units.

[[fig:0]]

When one vertex is the origin, the formula shrinks to 12 |x₁y₂ − x₂y₁|. And you can always make one vertex the origin by subtracting the coordinates of A from all three points. For the triangle above, B − A = (4, 1) and C − A = (2, 5), so the area = 12 |4 × 5 − 2 × 1| = 12 × 18 = 9. The answer is the same, with fewer steps.

[[fig:1]]

Area of a quadrilateral (or any polygon). Write the vertices in order around the figure (clockwise or anticlockwise, never criss-cross) and repeat the first one at the end. Multiply along the downward diagonals and along the upward diagonals, subtract, and halve:

Formula (Area of a quadrilateral (shoelace)):
>> Area = 12 |(x₁y₂ + x₂y₃ + x₃y₄ + x₄y₁) − (y₁x₂ + y₂x₃ + y₃x₄ + y₄x₁)|

This is called the shoelace method. You can also cut the quadrilateral ABCD along the diagonal AC into two triangles and add their areas.

[[fig:2]]

Collinearity. Three points lie on one straight line exactly when the "triangle" they make is flat, that is, its area is 0. An equivalent test: the slope from A to B equals the slope from B to C. Are (1, 4), (3, 8) and (5, 12) collinear? From the first point to the second, x goes up by 2 and y by 4. From the second to the third, x again goes up by 2 and y by 4. The steps are the same, so the points are collinear.

[[fig:3]]

When the area is given and a coordinate is unknown. Write the area formula with the unknown in it, set it equal to the given area, and solve. Because of the modulus sign, |expression| = A gives two equations, expression = 2A and expression = −2A, so there are usually two answers. Keep both unless the question fixes a sign or a range. Collinearity is the special case area = 0, which gives just one equation.

For Radhika's points (0, 0), (4, 0) and (2, k), the area is 12 × 4 × |k| = 2|k|. An area of 6 gives k = 3 or k = −3.

Now we come to the straight line, and the first thing to understand about any line is its slope (also called the gradient). The slope tells you how steep the line is: how much y rises when x increases by 1 unit to the right.

From two points. Go from A(x₁, y₁) to B(x₂, y₂). The horizontal change is the run (x₂ − x₁) and the vertical change is the rise (y₂ − y₁).

Formula (Slope):
>> m = riserun = y₂ − y₁x₂ − x₁

If the line makes an angle θ with the positive x-axis, then m = tan θ. You will study tan in Trigonometry; for now remember tan 45° = 1, tan 60° = √3 and tan 30° = 1√3.

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Radhika's cycle path goes from (1, 1) to (6, 4). The rise is 3 and the run is 5, so the slope is 35: for every 5 units forward, the path climbs 3 units.

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What the sign tells you:

• m > 0: the line goes up from left to right.
• m < 0: the line goes down from left to right.
• m = 0: a horizontal line (y = b).
• A vertical line (x = a) has run = 0, so its slope is not defined.

[[fig:2]]

From the equation.

• If the line is written as y = mx + c, the slope is simply the coefficient of x (when the coefficient of y is 1). For y = −5x + 2, m = −5.
• If the line is written as ax + by + c = 0 (or ax + by = c), make y alone: y = −abx + … So m = −ab = −coefficient of xcoefficient of y. For 4x − 6y = 7, m = −4−6 = 23.
• If the line cuts the axes at (p, 0) and (0, q), then m = q − 00 − p = −qp.

[[fig:3]]

Angle between two lines. If the slopes are m₁ and m₂, the acute angle θ between the lines satisfies tan θ = |m₁ − m₂1 + m₁m₂|. Equal slopes give θ = 0 (parallel lines). If m₁m₂ = −1, the denominator becomes 0, which means θ = 90°. These two special cases are the subject of Type 9.

[[fig:4]]

A straight line is fixed completely by two pieces of information: either a point and the slope, or two points. Every form of the equation is only a way of writing that information.

1. Slope-intercept form: y = mx + c. Here m is the slope and c is the y-intercept, the y-value where the line cuts the y-axis (put x = 0 and you get y = c). For y = 3x − 4 the slope is 3, and the line cuts the y-axis at (0, −4). If c = 0, the line y = mx passes through the origin.

[[fig:0]]

2. Point-slope form: y − y₁ = m(x − x₁). Use it when a point (x₁, y₁) and the slope m are given. Why does it work? For any other point (x, y) on the line, the slope from (x₁, y₁) to (x, y) must be m, that is, y − y₁x − x₁ = m. Cross-multiply and you have the form.

Radhika needs the line through (2, −1) with slope 3. She writes y + 1 = 3(x − 2), which gives y = 3x − 7.

[[fig:1]]

3. Two-point form. Given two points, first find the slope m = y₂ − y₁x₂ − x₁, then use the point-slope form with either point. Through (1, 4) and (3, 10): m = 62 = 3, and y − 4 = 3(x − 1) gives y = 3x + 1. The other point leads to the same line: y − 10 = 3(x − 3) ⇒ y = 3x + 1.

[[fig:2]]

4. General form: ax + by + c = 0. Every line can be written this way. To move from y = 25x + 3 to the general form, multiply by 5: 5y = 2x + 15 ⇒ 2x − 5y + 15 = 0. Options are often written in this form with whole numbers, so be ready to clear fractions and decimals.

Finding an unknown in the equation. If the line y = mx + 5 passes through (2, 11), the point must satisfy it: 11 = 2m + 5 ⇒ m = 3. The same idea finds c, or any other letter.

The y-intercept of a line through two points: find m first, then c = y₁ − mx₁.

The x-intercept of a line is the x-value where it cuts the x-axis, and the y-intercept is the y-value where it cuts the y-axis. You already know the method from the foundation:

• On the x-axis y = 0, so for the x-intercept put y = 0.
• On the y-axis x = 0, so for the y-intercept put x = 0.

For ax + by = c this gives x-intercept = ca and y-intercept = cb. For 5x − 4y = 20: put y = 0 ⇒ x = 4; put x = 0 ⇒ y = −5. So the line cuts the axes at (4, 0) and (0, −5).

[[fig:0]]

Read the options carefully. Some questions want the intercept as a number (4), others want the point (4, 0). An x-intercept point always has y = 0, so (0, 4) can never be an x-intercept point.

Intercept form. If a line cuts the x-axis at (a, 0) and the y-axis at (0, b), its equation is

Formula (Intercept form):
>> xa + yb = 1

Check: (a, 0) gives 1 + 0 = 1, and (0, b) gives 0 + 1 = 1. Suppose Radhika's line cuts the axes at 5 and −3. Its equation is x5 + y−3 = 1; multiplying by 15 gives 3x − 5y = 15.
To go the other way, divide by the constant: 4x + 3y = 12 ⇒ x3 + y4 = 1, so the intercepts are 3 and 4.

The triangle the line cuts off with the axes. The line, the x-axis and the y-axis make a right triangle with legs |a| and |b| and the right angle at the origin. So:

• Length of the portion between the axes = √a² + b² (it is the hypotenuse).
• Area of the triangle with the axes = 12 |a| × |b|. For a line written as Ax + By + C = 0 this becomes C²2|AB|.

For 4x + 3y = 12 the intercepts are 3 and 4, the length between the axes is 5 (the 3–4–5 triplet) and the area is 12 × 3 × 4 = 6.

[[fig:1]]

A related figure. The graph of |x| + |y| = k is a square standing on one corner, with corners (±k, 0) and (0, ±k). Each diagonal is 2k long, and the area is 2k².

[[fig:2]]

Two lines in a plane can relate in three ways: they are parallel, they cross at a right angle, or they cross at some other angle. The slope decides which.

Parallel lines have equal slopes: m₁ = m₂. They are equally steep and never meet.

• For a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, parallel means −a₁b₁ = −a₂b₂, that is, a₁a₂ = b₁b₂. In plain words, the x and y coefficients are in the same ratio and only the constant changes.
• So a line parallel to ax + by + c = 0 is ax + by + k = 0. Keep the left side, change the constant, and find k from the given point.

For example, the line through (1, 4) parallel to 2x − 5y = 3 is 2x − 5y = k, with k = 2 − 20 = −18. So the line is 2x − 5y = −18.

[[fig:0]]

Perpendicular lines have slopes whose product is −1: m₁ × m₂ = −1. So m₂ = −1m₁: turn the slope upside down and change its sign. Slope 3 gives the perpendicular slope −13; slope −25 gives 52.

• A line perpendicular to ax + by + c = 0 is bx − ay + k = 0. Swap the coefficients of x and y, change one sign, and then find k from the point.
• A special case: a horizontal line (slope 0) and a vertical line (slope not defined) are perpendicular, although the product rule cannot be applied to them.

Radhika needs the line through (2, 3) perpendicular to 4x + y = 9. Swapping the coefficients and changing one sign gives x − 4y = k, with k = 2 − 12 = −10. So the line is x − 4y = −10. Check: the slopes are −4 and 14, and their product is −1 ✓.

[[fig:1]]

Perpendicular bisector of AB. This is the line through the midpoint of AB that is perpendicular to AB. So there are two steps:
1. Find the midpoint M of AB.
2. Find the slope of AB, then turn it upside down and change its sign.
Then write the point-slope form through M. Every point on the perpendicular bisector is equidistant from A and B, and that is why the circumcentre lies on it.

[[fig:2]]

The point where two lines cross lies on both lines, so its coordinates satisfy both equations at the same time. Finding it is the same work as solving two linear equations in two variables, which you did in the Algebra chapter. Use whichever method is quicker:

• Elimination: add or subtract the equations to remove one variable. For x + 2y = 8 and x − y = 2, subtract: 3y = 6 ⇒ y = 2, then x = 4. The lines meet at (4, 2).
• Substitution or equating: if both are in the form y = …, equate them. For y = 3x − 1 and y = x + 5: 3x − 1 = x + 5 ⇒ x = 3, y = 8.

In the exam, checking the options is often the fastest way: the correct point must satisfy both equations.

[[fig:0]]

How can two lines meet? For a₁x + b₁y = c₁ and a₂x + b₂y = c₂, compare the ratios:

Take 2x − 3y = 4 and 6x − 9y = 12. The ratios are 26 = −3−9 = 412 = 13, all equal. The second equation is just 3 times the first, so the two lines coincide. Two straight lines can never meet at exactly two points.

[[fig:1]]

Using the intersection. Many questions use the intersection point as a middle step:

• "The intersection of these two lines lies on a third line; find k." Find the point, then put it in the third line. Three lines through one point are called concurrent.
• "Find the area bounded by two lines and the x-axis." The region is a triangle. Its base is the gap between the two x-intercepts and its height is the y-coordinate of the intersection point. In the figure, x + y = 6 and x − 2y = 0 meet at P(4, 2) and cut the x-axis at 6 and 0, so the area = 12 × 6 × 2 = 6.

[[fig:2]]

Distance of a point from a line. There are endless paths from a point P to a line, but the shortest one is the perpendicular, and its length is the distance of P from the line. For P(x₁, y₁) and the line ax + by + c = 0:

Formula (Distance of a point from a line):
>> d = |ax₁ + by₁ + c|√a² + b²

Bring the equation to the form "everything = 0", put the point into the left side, take the absolute value, and divide by √a² + b².

For example, the distance of P(5, 4) from 3x + 4y − 12 = 0 is |15 + 16 − 12|√9 + 16 = 195 = 3.8.

[[fig:0]]

From the origin the formula becomes d = |c|√a² + b². There is a neat picture behind it. The line cuts the axes at p and q, and with the origin it forms a right triangle. The perpendicular from the origin is the altitude on the hypotenuse, and altitude = product of the legshypotenuse = |pq|√p² + q² (take the intercepts without their signs). For 3x + 4y = 60 the intercepts are 20 and 15, the hypotenuse is 25, and the distance from O is 30025 = 12. No formula is needed.

[[fig:1]]

Distance between two parallel lines ax + by + c₁ = 0 and ax + by + c₂ = 0 (the same a and b): d = |c₁ − c₂|√a² + b².

[[fig:2]]

The circle. A circle is the set of all points at a fixed distance r (the radius) from a fixed point (h, k) (the centre). Write the distance formula for a general point (x, y):

Formula (Circle):
>> (x − h)² + (y − k)² = r²

• Centre at the origin: x² + y² = r². So x² + y² = 49 is a circle of radius 7.
• Expanded form x² + y² + 2gx + 2fy + c = 0: centre (−g, −f), radius √g² + f² − c.

Radhika's circle has centre (2, −1) and radius 3: (x − 2)² + (y + 1)² = 9, which expands to x² + y² − 4x + 2y − 4 = 0.

[[fig:3]]

Circle touching a line. If a line is a tangent to a circle, the radius drawn to the point of contact is perpendicular to the line. So radius = perpendicular distance of the centre from the tangent.

[[fig:4]]

One step further: the sphere. In three dimensions a point has three coordinates (x, y, z), and the same distance idea gives the sphere with centre (a, b, c) and radius r: (x − a)² + (y − b)² + (z − c)² = r².

Writing the circle from the given conditions. Every circle question ends with the same form (x − h)² + (y − k)² = r²; the work is only in finding the centre and the radius.

• Ends of a diameter (x₁, y₁) and (x₂, y₂): the centre is their midpoint and r is half the diameter. Faster, the diameter form (x − x₁)(x − x₂) + (y − y₁)(y − y₂) = 0 gives the equation in one line.
• Centre and a point on the circle: r is the distance between them.
• Touches the x-axis: r = |k|. Touches the y-axis: r = |h|. Touches both axes: the centre is (±r, ±r), so the equation is (x ∓ r)² + (y ∓ r)² = r²; in the first quadrant the centre is (r, r).

[[fig:5]]

For Radhika's diameter from (1, 1) to (5, 5), the diameter form gives (x − 1)(x − 5) + (y − 1)(y − 5) = 0, that is, x² + y² − 6x − 6y + 10 = 0, with centre (3, 3) and radius √8.

In Type 10 you found the meeting point of two lines and then used it. Many questions want a line through that meeting point with one extra condition: it also passes through some point, or it is parallel or perpendicular to a given line, or it cuts equal intercepts. There are two ways to do this, and both are short.

Way 1: find the point first. Solve the two equations, get the point P, and then write the required line through P with the condition. For x + y = 6 and 2x − y = 0, adding gives 3x = 6, so P = (2, 4). The line through P parallel to the x-axis is y = 4.

Way 2: the family L₁ + λL₂ = 0. Write both lines in the form "… = 0": L₁ = x + y − 6 and L₂ = 2x − y. For any number λ, the equation L₁ + λL₂ = 0 is a straight line through the meeting point of L₁ and L₂, because at P both L₁ and L₂ are zero. Changing λ turns the line around P. Use the extra condition to find λ. This saves you from solving the two equations when the condition is "passes through a given point".

[[fig:0]]

Concurrent lines. Three lines are concurrent when they all pass through one point. The test is simple: find the meeting point of any two of them and put it in the third. Radhika checks x + y = 6, 2x − y = 0 and 3x − y = 2: the first two meet at (2, 4), and 3 × 2 − 4 = 2, so the third line also passes through it. The same condition in determinant form, for lines a_ix + b_iy + c_i = 0, is
| a₁ b₁ c₁ | a₂ b₂ c₂ | a₃ b₃ c₃ | = 0.
In the exam the substitution method is quicker.

[[fig:1]]

A useful special case. For the family L₁ + λL₂ = 0, the member parallel to the y-axis is the one in which the coefficient of y becomes zero, and the member parallel to the x-axis is the one in which the coefficient of x becomes zero.

Type 1 gave you the mirror images in the axes, the origin and the lines y = ±x. This type completes the picture with three more moves: translation, rotation and reflection in any line. In every one of them the shape does not change; only its position does. So lengths, angles and areas stay the same.

Translation. Sliding a point a units to the right and b units up sends (x, y) to (x + a, y + b). Left and down simply make a or b negative. Radhika moves (2, 5) three units left and one unit down and lands at (−1, 4).

Rotation about the origin. Turning through 90° anticlockwise sends (x, y) → (−y, x): the coordinates swap and the new first coordinate changes sign. Through 180° the point goes to (−x, −y), which is the same as reflection in the origin. Through 90° clockwise (or 270° anticlockwise), (x, y) → (y, −x).

[[fig:0]]

To rotate about some other point (h, k), shift that point to the origin, rotate, and shift back. In symbols, move to (x − h, y − k), apply the rule, then add (h, k) again.

Reflection in x = a and y = b. The mirror x = a is a vertical line. The image is as far on the other side, so the x-coordinate becomes 2a − x and y stays: (x, y) → (2a − x, y). In the mirror y = b, (x, y) → (x, 2b − y). Take P(1, 3) and the mirror x = 3: the image is (6 − 1, 3) = (5, 3).

[[fig:1]]

Reflection in any line ax + by + c = 0. Put the point in the line and divide by a² + b²: k = ax₁ + by₁ + ca² + b². Then

• the foot of the perpendicular is (x₁ − ka, y₁ − kb), and
• the image is (x₁ − 2ka, y₁ − 2kb), because the foot is the midpoint of the point and its image.

For the origin and the line x + y − 2 = 0, k = −22 = −1, so the foot is (1, 1) and the image is (2, 2).

[[fig:2]]

To reflect a whole line in a mirror, replace each coordinate by the coordinate of its image. For example, in the mirror x = a replace x by 2a − x in the equation.