--- title: "Parallel and Perpendicular Lines in Coordinates" description: "Use equal gradients for parallel lines and a gradient product of -1 for perpendicular ones, and apply both to find missing lines, midpoints and distances." canonical: https://lightmysky.com/learn/mathematics/parallel-and-perpendicular-lines-in-coordinates-mt_xNEmm139Ka source: https://lightmysky.com/learn/mathematics/parallel-and-perpendicular-lines-in-coordinates-mt_xNEmm139Ka.md retrieved: 2026-09-02 --- > **Agent view.** This is the Markdown twin of the page, for tools and assistants. > When to use this site, and the call that answers each job: https://lightmysky.com/agent-instructions.md > API description (OpenAPI 3.1): https://lightmysky.com/openapi.json · Authentication: https://lightmysky.com/auth.md > Pricing: https://lightmysky.com/pricing.md · Catalog: https://lightmysky.com/llms.txt · Full catalog: https://lightmysky.com/llms-full.txt > Every machine-readable file on this domain: https://lightmysky.com/.well-known/ai-catalog.json > Ask for Markdown with `Accept: text/markdown`, a `.md` address, or `?mode=agent`. # Parallel and Perpendicular Lines in Coordinates Use equal gradients for parallel lines and a gradient product of -1 for perpendicular ones, and apply both to find missing lines, midpoints and distances. Subject: Mathematics · Area: Geometry · Ages 16 to 17 Page: https://lightmysky.com/learn/mathematics/parallel-and-perpendicular-lines-in-coordinates-mt_xNEmm139Ka ## Ready when they can - Find the line through (1, 4) perpendicular to y = 2x - 3 - Show that two given lines are parallel and never meet - Find the perpendicular bisector of the segment joining two points ## Lesson: Parallel and Perpendicular Lines in Coordinates Last stop gave you the equation of any one line. The Ridgeway practice strip needs two at a time: its long edges have to stay the same distance apart, and the gate path has to meet them square. Both jobs are settled by gradients alone. Two lines are parallel when their gradients are equal and their intercepts are not. If both match, they are not two lines at all, they are one line written twice. Neither test works until both equations are in y = mx + c, so rearrange first and compare second. Perpendicular is the other relation, and its test is a product. Two lines meet at a right angle exactly when their gradients multiply to -1. That makes each gradient the negative reciprocal of the other: flip the fraction and change the sign. A gradient of 3 pairs with -1/3, and -3/4 pairs with 4/3. Whole numbers are fractions too, so 6 is 6/1 and pairs with -1/6. The quickest check on any pair is to multiply them, not to eyeball the signs, because two gradients can have opposite signs and still cross at an ordinary angle. *(drawing: Three outcomes from comparing two gradients. Rearrange into y = mx + c first.)* **Example.** The gate path runs through the peg at (1, 4) and has to meet the edge y = 2x - 3 square. The edge's gradient is 2, so the path's is -1/2. From there it is the previous stop's job: y - 4 = -(1/2)(x - 1). Multiply every term by 2 to clear the fraction: 2y - 8 = -x + 1. Moving everything to one side gives x + 2y - 9 = 0. Check the point: 1 + 8 - 9 = 0. Check the gradient: 2y = -x + 9 gives y = -(1/2)x + 4.5, and 2 times -1/2 is -1. Two more tools come off the same coordinates. The midpoint of a segment is the average of the two x values and the average of the two y values, so (2, 1) and (14, 10) have midpoint (8, 5.5). The length is Pythagoras on the differences: the horizontal gap is 14 - 2 = 12, the vertical gap is 10 - 1 = 9, and the segment is the hypotenuse, so its length is the square root of 12² + 9², which is 15. Differences can come out negative, but squaring removes the sign, so the order does not matter. *(drawing: Average for a point, Pythagoras for a length. The bisector takes the midpoint and the flipped gradient.)* **Example.** A perpendicular bisector halves a segment at a right angle, so it needs a midpoint and a gradient. Take (1, 2) and (5, 8), midpoint (3, 5). The segment's gradient is (8 - 2) / (5 - 1) = 3/2, so the bisector's is -2/3. Then y - 5 = -(2/3)(x - 3), and multiplying by 3 gives 2x + 3y - 21 = 0. Check the midpoint: 6 + 15 - 21 = 0. A flat segment has gradient 0, which cannot be flipped, so its bisector is the vertical line x = a through its midpoint. Every point on a bisector is the same distance from both ends. **Recap.** Two lines are parallel when their gradients are equal, and perpendicular when their gradients multiply to -1, which makes each the negative reciprocal of the other. Rearrange into y = mx + c before comparing. The midpoint averages the two coordinates, and the length is Pythagoras on the differences. A perpendicular bisector uses both: the midpoint for a point, the negative reciprocal for a gradient. ## Practice 24 questions on this page, each with its working shown. ## Needs first - [Forms of the Equation of a Straight Line](https://lightmysky.com/learn/mathematics/forms-of-the-equation-of-a-straight-line-mt_dhIIxdFwcp) - [Pythagoras' Theorem](https://lightmysky.com/learn/mathematics/pythagoras-theorem-mt_1VmTUxBrNd) ## Opens up - [The Equation of a Circle](https://lightmysky.com/learn/mathematics/the-equation-of-a-circle-mt_YARqmi_ioe) - [Tangents and Normals to a Curve](https://lightmysky.com/learn/mathematics/tangents-and-normals-to-a-curve-mt_BzMw7IT6kp)