Coordinates: from a drawing to an equation

A plan gives the coordinates of two opposite points on a circular plot. Can we find the center and check whether a new marker lies on the boundary without measuring a screen with a ruler? We translate geometric conditions into equations.

Coordinates specify position; a figure’s equation selects all and only its points. Distance follows from Pythagoras, a circle from constant distance to a center, and a line from a linear relation. We use G014, G019 and linear functions.

Agree on axes first

Choose origin O, perpendicular x,y axes, positive directions and a length unit. In the pair (x;y), the first number is the horizontal coordinate and the second the vertical coordinate in this drawing. These are not distances from the origin: they can be negative. Points on the x-axis have y=0; points on the y-axis have x=0. The axes themselves belong to none of the four open quadrants.

For the ordinary distance formula, both coordinates use the same physical length unit. Draw equal unit lengths on both axes to see the geometry accurately. Stretching a graph horizontally can make a circle look oval: the equation has stayed the same, while the image is distorted. A time–cost graph does not directly represent Euclidean metres between points.

Distance is a hypotenuse

For A(x₁;y₁),B(x₂;y₂), a right triangle has legs |x₂−x₁| and |y₂−y₁|. Pythagoras gives:

$AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.$

Squaring already removes the signs of the differences. If one coordinate is shared, one leg vanishes and the formula gives distance along the other axis. For A=B, distance is 0. OpenStax, distance and midpoint.

Take our own A(−2;−5),B(4;3), coordinates in metres. Differences 6 and 8 give AB=√(36+64)=10 m. This is straight-line distance; travelling only along coordinate directions would cover 6+8=14 m.

Distance and midpoint

Circle and point checks

A vertical line

Coordinates in metres, equal scales on both axes. A(−2;−5), B(4;3), Q(1;−1). First: horizontal change 6, vertical change 8, AB=10, Q bisects AB. Second: circle centered at Q with radius 5; B on it, P(5;3) outside, T(3;1) inside. The solid sloping line is the visible part of 4x−3y−7=0. Third: x=1, every point has the same x and Δx=0. Axis labels are coordinates, not ordinal numbers.

Midpoint: average each coordinate

The midpoint makes half the horizontal and half the vertical change. Its coordinates are:

$M\left(\frac{x_1+x_2}{2};\frac{y_1+y_2}{2}\right).$

For A,B we get Q(1;−1). From A to Q the changes are 3 and 4; from Q to B they are also 3 and 4. Thus AQ=QB=5. Do not average all four numbers together: x and y describe different directions.

A circle is a constant-distance condition

Let AB be a diameter. Center Q(1;−1), radius r=5. Point X(x;y) lies on the circle if and only if QX=5. Squaring gives:

$(x-1)^2+(y+1)^2=25.$

For a general center Q(a;b) and r>0, this is (x−a)²+(y−b)²=r². The sign inside each bracket is opposite to the center’s coordinate: y+1 means y−(−1). The right side is the squared radius. UC Davis, rectangular coordinates.

Substituting B(4;3) gives 9+16=25: it is on the boundary. For P(5;3), 16+16=32>25, outside the disk. For T(3;1), 4+4=8<25, inside it. The inequality ≤25 describes the disk; equality to 25 describes the circle. One function y(x) cannot describe the entire circle: many x values have two y values.

A line and its slope

When x₂≠x₁, slope k=(y₂−y₁)/(x₂−x₁). For AB it is 8/6=4/3. Using A:

$y+5=\frac43(x+2),\qquad 4x-3y-7=0.$

Both forms describe the same line; substitute A,B to check the numbers. With matching units and scales, k is the tangent of the inclination angle measured from positive x, provided the line is not vertical. A decreasing line has negative slope; a horizontal line has slope 0. OpenStax, slope.

A vertical line x=1 has Δx=0: division by zero is undefined. Its slope is undefined, rather than some infinite real number. It remains a line. The general form ax+by+c=0 includes it, provided a,b are not both zero. If a=b=0, we get the whole plane when c=0 or no points when c≠0, rather than a line.

The coordinate methodOptional · safe to skip

Equal distances give a perpendicular bisector

Points equally distant from A and B satisfy (x+2)²+(y+5)²=(x−4)²+(y−3)². Expanding cancels the quadratic terms: 12x+16y+4=0, or 3x+4y+1=0. This is AB’s perpendicular bisector, as in G009. It passes through Q(1;−1); slope −3/4 agrees with perpendicularity to a line of slope 4/3.

Finish a coordinate solution by returning to the figure: which points did the equation select, what units were used, and was the vertical case preserved? Checking the numbers then checks a geometric condition.