Coordinate Geometry Practice Questions

Use the distance formula:

\(d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}\)

Substitute the given points:

\(d = \sqrt{(6 – 2)^2 + (6 – 3)^2}\)\(\:= \sqrt{16 + 9} = \sqrt{25} = 5\)

Choice D gives \((x_2 – x_1)^2 + (y_2 – y_1)^2 = 25\) without taking the square root. Remember: the distance formula always ends with a square root.

The midpoint formula finds the point exactly halfway between two given points by averaging the \(x\)-coordinates and the \(y\)-coordinates separately:

\(M = \left(\dfrac{x_1 + x_2}{2},\: \dfrac{y_1 + y_2}{2}\right)\)

Substitute:

\(M = \left(\dfrac{1 + 5}{2},\: \dfrac{7 + 3}{2}\right) = (3, 5)\)

Choice B adds the coordinates without dividing by 2. Choice C subtracts the coordinates instead of adding them.

Let the unknown endpoint be \((x, y)\). Since the midpoint formula averages the endpoints, set up two equations:

\(\dfrac{4 + x}{2} = 7 \implies 4 + x = 14 \implies x = 10\)

\(\dfrac{1 + y}{2} = 5 \implies 1 + y = 10 \implies y = 9\)

The other endpoint is \((10, 9)\).

Choice B gives the midpoint between \((4, 1)\) and \((7, 5)\) — that’s the midpoint of the wrong segment. To find a missing endpoint, multiply the midpoint coordinates by 2 and subtract the known endpoint.

Points are collinear if they all lie on the same line. To test this, check whether the slope between each pair of consecutive points is the same:

Slope from \((1, 2)\) to \((3, 6)\): \(\frac{6 – 2}{3 – 1} = \frac{4}{2} = 2\)

Slope from \((3, 6)\) to \((5, 10)\): \(\frac{10 – 6}{5 – 3} = \frac{4}{2} = 2\)

Both slopes equal 2, so all three points lie on the same line and are collinear.

Choice C uses the wrong test. Equal distances between consecutive points means the points are equally spaced, but they could still form a zigzag rather than a straight line.

Find the length of each side using the distance formula:

• \(AB = \sqrt{(4-0)^2 + (0-0)^2} = 4\)
• \(AC = \sqrt{(0-0)^2 + (3-0)^2} = 3\)
• \(BC = \sqrt{(4-0)^2 + (0-3)^2} = \sqrt{16+9} = 5\)

Check using the Pythagorean theorem: if \(a^2 + b^2 = c^2\), the triangle is a right triangle:

\(3^2 + 4^2 = 9 + 16 = 25 = 5^2\) ✓

This is a classic 3-4-5 right triangle with the right angle at vertex A.

Notice that points A and B share the same \(y\)-coordinate (\(y = 2\)), so side AB is horizontal. This makes it easy to identify the base and height:

• Base (AB): \(|5 – 1| = 4\)
• Height: The vertical distance from C to line AB is \(|6 – 2| = 4\)

Apply the triangle area formula:

\(A = \dfrac{1}{2}bh = \dfrac{1}{2}(4)(4) = 8\)

When vertices share a coordinate value, look for a horizontal or vertical side to use as the base — it simplifies the calculation greatly.

Find the length of each side using the distance formula:

• \(AB = \sqrt{(6-0)^2 + (0-0)^2} = 6\)
• \(BC = \sqrt{(6-3)^2 + (0-4)^2} = \sqrt{9+16} = 5\)
• \(AC = \sqrt{(3-0)^2 + (4-0)^2} = \sqrt{9+16} = 5\)

Add all three sides:

\(P = 6 + 5 + 5 = 16\)

This is an isosceles triangle since two sides (BC and AC) have equal length. Notice that the perimeter requires the distance formula for each side, not just the differences in coordinates.

When a point divides a segment in the ratio \(m:n\), use the section formula:

\(P = \left(\dfrac{m \cdot x_2 + n \cdot x_1}{m + n},\: \dfrac{m \cdot y_2 + n \cdot y_1}{m + n}\right)\)

With \(A(1, 3)\), \(B(7, 9)\), and ratio 2:1:

\(x = \dfrac{2(7) + 1(1)}{2 + 1} = \dfrac{15}{3} = 5\)

\(y = \dfrac{2(9) + 1(3)}{2 + 1} = \dfrac{21}{3} = 7\)

So \(P = (5, 7)\). Note that a 2:1 ratio means P is \(\frac{2}{3}\) of the way from A to B, which is closer to B than to A.

Choice B gives the midpoint (ratio 1:1), not the 2:1 division point.

The distance from a point \((x_0, y_0)\) to a line \(Ax + By + C = 0\) is:

\(d = \dfrac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}\)

Substitute \((x_0, y_0) = (3, 4)\) and \(A = 3\), \(B = 4\), \(C = -5\):

\(d = \dfrac{|3(3) + 4(4) – 5|}{\sqrt{3^2 + 4^2}} = \dfrac{|9 + 16 – 5|}{\sqrt{9 + 16}}\)\(\:= \dfrac{20}{5} = 4\)

Choice C gives only the numerator (20) without dividing by \(\sqrt{A^2 + B^2}\). The absolute value in the numerator ensures the distance is always positive.

A perpendicular bisector passes through the midpoint of the segment and has a slope that is the negative reciprocal of the segment’s slope.

Step 1

Find the midpoint:

\(M = \left(\dfrac{2+6}{2},\: \dfrac{4+8}{2}\right) = (4, 6)\)

Step 2

Find the slope of the segment:

\(m = \dfrac{8-4}{6-2} = \dfrac{4}{4} = 1\)

Step 3

The perpendicular slope is −1. Use point-slope form with the midpoint:

\(y – 6 = -1(x – 4) \implies y = -x + 10\)

A point lies on a line if its coordinates satisfy the equation. Substitute each point’s \(x\)-value into \(y = 3x + 1\) and check if the result matches the \(y\)-value:

• Choice A: \(3(2) + 1 = 7\). The point is \((2, 7)\) ✓
• Choice B: \(3(3) + 1 = 10 \neq 8\) ✗
• Choice C: \(3(1) + 1 = 4 \neq 5\) ✗
• Choice D: \(3(4) + 1 = 13 \neq 11\) ✗

Only \((2, 7)\) satisfies the equation, so it is the only point that lies on the line.

The center of the circle is the midpoint of the diameter:

\(\text{center} = \left(\dfrac{2+8}{2},\: \dfrac{3+11}{2}\right) = (5, 7)\)

The radius is half the length of the diameter. First, find the diameter length:

\(d = \sqrt{(8-2)^2 + (11-3)^2}\)\(\:= \sqrt{36+64} = \sqrt{100} = 10\)

So \(r = \frac{10}{2} = 5\), and the equation is:

\((x – 5)^2 + (y – 7)^2 = 25\)

Choice B uses the diameter (10) instead of the radius (5) when computing \(r^2\).

When a point is reflected over the \(x\)-axis, the \(x\)-coordinate stays the same and the \(y\)-coordinate changes sign:

\((x, y) \rightarrow (x, -y)\)

So \((3, -5)\) becomes \((3, 5)\).

For reference, the reflection rules are:

• Over the \(x\)-axis: \((x, y) \rightarrow (x, -y)\) — negate \(y\)
• Over the \(y\)-axis: \((x, y) \rightarrow (-x, y)\) — negate \(x\)
• Over the origin: \((x, y) \rightarrow (-x, -y)\) — negate both

Choice B reflects over the \(y\)-axis, and choice C reflects over the origin.

Apply the distance formula:

\(d = \sqrt{(9-3)^2 + (15-7)^2}\)\(\:= \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\)

Since each unit on the grid represents 1 mile, the towers are 10 miles apart.

Choice B adds the horizontal and vertical distances (\(6 + 8 = 14\)), which gives the “walking distance” along a grid rather than the straight-line distance. The distance formula computes the straight-line (“as the crow flies”) distance.

From the coordinates, identify the key measurements. The two parallel sides (bases) are horizontal:

• Base 1 (AB): \(|8 – 0| = 8\) (along \(y = 0\))
• Base 2 (DC): \(|6 – 2| = 4\) (along \(y = 5\))
• Height: \(|5 – 0| = 5\) (vertical distance between the two bases)

Apply the trapezoid area formula:

\(A = \dfrac{1}{2}(b_1 + b_2)(h) = \dfrac{1}{2}(8 + 4)(5)\)\(\:= \dfrac{1}{2}(60) = 30\)

Coordinate geometry makes it easy to find lengths and heights when sides are horizontal or vertical — just subtract the relevant coordinates.