Reference y = 2x + 1, point (3, −2)
The parallel line keeps slope 2: y+2=2(x−3), so y=2x−8. The perpendicular slope is −1/2: y+2=−1/2(x−3), or x+2y=−1.
Enter a reference line and a point. Get the unique parallel and perpendicular lines through that point in exact standard, point-slope, and graphing forms, with substituted steps and a graph.
Try an example:
Writing the source line as Ax + By = C gives one method that works for every orientation, including vertical and horizontal lines. Let the required point be P=(x₀,y₀).
Keep the same normal vector (A,B) and change the constant so the new point satisfies the equation:
Ax + By = Ax₀ + By₀
Rotate the normal vector by 90°. One valid perpendicular normal is (B,−A):
Bx − Ay = Bx₀ − Ay₀
For a finite, nonzero source slope m, the new slopes are:
mparallel=m, mperpendicular=−1/m
A horizontal source line has slope 0, so its perpendicular is vertical rather than having a finite negative-reciprocal slope. A vertical source has undefined slope, and its perpendicular is horizontal. Standard form handles both without division.
(x₀,y₀) that both new lines must pass through.The parallel line keeps slope 2: y+2=2(x−3), so y=2x−8. The perpendicular slope is −1/2: y+2=−1/2(x−3), or x+2y=−1.
The parallel line is vertical through the new x-coordinate, x=−2. Its perpendicular is horizontal through the new y-coordinate, y=5.
The parallel line is 3x−2y=10. A rotated normal is (−2,−3), giving the equivalent reduced perpendicular equation 2x+3y=11.
Ax+By=C represents every line, including vertical lines. The calculator clears denominators and reduces common integer factors.
y−y₀=m(x−x₀) makes the required point and slope visible. Vertical lines are shown as x=x₀.
If the required point lies on the reference line, the parallel result is the reference line itself. This is reported explicitly rather than called a distinct parallel.
Fraction inputs remain rational throughout the line calculations. Decimal approximations are added only where they help interpretation.
m=2 has slope −1/2, not −2.−1/m when m=0; the perpendicular is vertical.A=0 and B=0 in standard form, which does not define a line.For a source Ax+By=C and point (x₀,y₀), keep A and B and use Ax+By=Ax₀+By₀.
Rotate the source normal (A,B) to (B,−A). The line through (x₀,y₀) is Bx−Ay=Bx₀−Ay₀.
When m is finite and nonzero, the perpendicular slope is −1/m. Horizontal and vertical lines are the special pair.
The parallel result coincides with the source line. The perpendicular result remains unique.
Yes. Use standard form such as 1x+0y=4, or use two points with the same x-coordinate. The outputs automatically switch to x=k or y=k where appropriate.
Yes. Examples include -3/4, -2 1/3, 1.25, and 4e-3. Exact rational equations are preserved.
Calculation method: inputs are converted to reduced rational numbers. Source, parallel, and perpendicular equations are calculated with integer arithmetic, denominators are cleared, common factors are removed, and signs are normalized. Only the requested decimal values and graph use floating-point arithmetic.
Formula reference: the same-slope and negative-reciprocal rules, point-slope construction, and horizontal/vertical cases follow OpenStax College Algebra 2e, “Linear Functions”. The calculator extends these rules to standard-form vector equations so no slope division is needed.
Verification completed July 30, 2026. Checks cover general, standard-form, two-point, fractional, horizontal, vertical, coincident, empty, malformed, degenerate, and large-value cases.