Lesson 82 of 100 | Quantitative Aptitude / Algebra / बीजगणित
Graphs of linear equations
Learning outcome
Describe lines from coordinate tables and interpret their intersections.
Concepts and assumptions
Use real coordinates. The horizontal x-axis and vertical y-axis meet perpendicularly at (0, 0). Positive directions are right and up; negative directions are left and down. Use equal one-unit intervals on both axes.
In (x, y), horizontal position comes first. A graph contains every allowed solution pair, not just tabulated samples.
For ax + by = c, a and b are not both zero. If b ≠ 0, then y = −(a/b)x + c/b. Slope measures vertical change divided by nonzero horizontal change. If b = 0, a ≠ 0: x = c/a is vertical. A line y = k is horizontal.
Two distinct points determine a line. Extend both ways: domains here are unrestricted. At x-axis intercepts y = 0; at y-axis intercepts x = 0. Intersections satisfy both equations. Distinct parallel lines have none; coincident lines share infinitely many points.
Worked examples
Example 1 — Descending line. Describe x + 2y = 6.
Rearrange: y = (6 − x)/2.
| x | 0 | 2 | 6 |
|---|---|---|---|
| y | 3 | 2 | 0 |
For x = 2, y = (6 − 2)/2 = 2. Join the points. It falls one unit per two units rightward, with intercepts (0, 3) and (6, 0).
Example 2 — Intersection. Compare x + y = 5 and x − y = 1.
Rewrite as y = 5 − x and y = x − 1.
| x | 0 | 3 | 5 |
|---|---|---|---|
| y = 5 − x | 5 | 2 | 0 |
| y = x − 1 | −1 | 2 | 4 |
One descends; the other rises. Equating gives 5 − x = x − 1, so x = 3 and y = 2. Intersection: (3, 2). Check: 3 + 2 = 5; 3 − 2 = 1.
Example 3 — Vertical line. Describe x = −2.
| x | −2 | −2 | −2 |
|---|---|---|---|
| y | −2 | 0 | 3 |
All points are two units left of the y-axis. The vertical line crosses only the x-axis, at (−2, 0). Its slope is undefined: horizontal change is zero.
Common mistakes
Reversing coordinates; uneven scales; drawing only a segment; assuming every line meets both axes; dividing by zero for vertical slope.
Practice questions
- Tabulate 2x + y = 4 at x = 0, 1, 2; state its intercepts.
- Give three points on y = −3; describe its line and intercepts.
- Find and verify the intersection of y = x + 1 and y = −x + 5.
- Compare y = 2x + 1 with y = 2x − 3, then with 2y = 4x + 2. Count intersections.
Worked solutions
- Rearrange to y = 4 − 2x:
| x | 0 | 1 | 2 |
|---|---|---|---|
| y | 4 | 2 | 0 |
The line descends. Intercepts: (0, 4) and (2, 0).
- Points (−2, −3), (0, −3), (2, −3) form a horizontal line three units below the x-axis. Its y-intercept is (0, −3); no x-intercept exists.
- Equate: x + 1 = −x + 5, so 2x = 4. Thus (x, y) = (2, 3). Check: 3 = 2 + 1 and 3 = −2 + 5.
- The first pair has slope 2 but different y-intercepts; equating gives 1 = −3: no intersection. Dividing 2y = 4x + 2 by 2 reproduces the first equation: infinitely many intersections.