Apply the distance, midpoint and slope formulas, identify parallel and perpendicular lines, and analyse figures placed in the coordinate plane (Geometry).

What this topic is asking

Coordinate geometry places points, lines and figures on the $xy$-plane and uses algebraic formulas to measure them. The ACT tests the distance, midpoint and slope formulas, parallel and perpendicular lines, and analysing figures (such as a triangle's sides) from their coordinates.

The three core formulas

These three formulas drive coordinate geometry.

The distance formula is just the Pythagorean theorem: the horizontal gap and vertical gap are the legs, and the distance is the hypotenuse.

Using distance and midpoint

The midpoint finds the centre of a segment, which is also the centre of a circle whose diameter is that segment.

Parallel and perpendicular lines

Slope relationships identify how two lines meet.

• Parallel lines never cross and have equal slopes.
• Perpendicular lines meet at a right angle and have negative-reciprocal slopes: $m_1 m_2 = -1$, so a line of slope $\frac{3}{4}$ is perpendicular to one of slope $-\frac{4}{3}$.

This lets you test, from coordinates alone, whether two segments form a right angle (compute both slopes and check their product is $-1$) or are parallel (equal slopes), which is how the ACT asks you to classify a quadrilateral or confirm a right triangle.

Analysing figures from coordinates

A frequent ACT task gives the vertices of a triangle or quadrilateral and asks for a side length, a perimeter, whether it has a right angle, or whether two sides are parallel. The method combines the three formulas: use distance for side lengths and perimeter, slope to test for parallel sides or right angles, and midpoint for diagonals' centres. For example, to show a triangle is right-angled, find the slopes of two sides and check their product is $-1$. Breaking the figure into these algebraic checks turns a geometry question into routine computation.

Horizontal and vertical segments

When two points share an $x$-coordinate or a $y$-coordinate, the distance is just the difference of the other coordinate, with no square root needed. Points $(3, 1)$ and $(3, 9)$ lie on a vertical segment of length $9 - 1 = 8$; points $(2, 5)$ and $(7, 5)$ lie on a horizontal segment of length $7 - 2 = 5$. Recognising these shortcuts is faster than the full distance formula and is common in ACT questions that build a rectangle or right triangle from grid points. It also helps compute the area of a figure whose sides are horizontal and vertical, since base and height are read straight off the coordinates.

Reflections and translations of points

The ACT sometimes moves a point in the plane. Reflecting across the $x$-axis negates the $y$-coordinate ($(a, b) \to (a, -b)$); across the $y$-axis negates the $x$-coordinate ($(a, b) \to (-a, b)$); across the origin negates both. A translation adds fixed amounts to the coordinates, so shifting a point right 3 and up 2 sends $(a, b)$ to $(a + 3, b + 2)$. These coordinate rules let you find the image of a vertex without re-drawing the figure, and they connect to the function-transformation rules in the Functions area.

Why these formulas connect algebra and geometry

Coordinate geometry is where algebra and shape meet: a side length becomes a distance computation, a right angle becomes a slope product, a centre becomes an average. Keeping the three formulas straight, and remembering the distance formula is the Pythagorean theorem, lets you answer figure questions by calculation rather than by eye. Careful subtraction order and not forgetting the square root or the halving are the habits that secure these points.

Try this

Q1. Find the distance between $(2, 1)$ and $(2, 9)$. [1 point]

• Cue. Same $x$, so it is the vertical gap: $\sqrt{0 + 8^{2}} = 8$.

Q2. Find the midpoint of $(1, 4)$ and $(7, 10)$. [1 point]

• Cue. $\left(\frac{1 + 7}{2}, \frac{4 + 10}{2}\right) = (4, 7)$.