6.2 Slope, Equations of Lines & Intercepts

Slope as a Rate

Slope is the constant rate of change of y with respect to x along a non-vertical line:

m = (y₂ − y₁) / (x₂ − x₁) = rise / run

Order of the two points does not matter: swapping both pairs of coordinates cancels the two minus signs. Mixing the order — using y₂ − y₁ over x₁ − x₂ — flips the sign and is a common miss.

Four pictures, four algebraic facts:

A vertical line through (4, −1) and (4, 9) has Δx = 0. The slope fraction is 10/0, which is undefined — not 0, and not a number you should write as "infinity" on Quant. Zero slope is the horizontal line y = constant. Mixing those two cases is one of the highest-yield coordinate traps in the section.

Worked example. Slope from (−3, 1) to (1, 9): m = (9 − 1)/(1 − (−3)) = 8/4 = 2.

Worked example (horizontal vs vertical). Through (5, 7) and (−1, 7): Δy = 0, so m = 0 and the equation is y = 7. Through (5, 7) and (5, −2): Δx = 0, slope undefined, equation x = 5.

Three distinct points A, B, and C are collinear when slope AB equals slope BC (and therefore AC). If the slopes disagree, the points are not on one line. Collinearity is an equal-rate test, not a triangle-angle theorem.

Three Forms of a Line

Horizontal: y = c (slope 0). Vertical: x = k (undefined slope). A vertical line is not a function of x; every other line in the table is.

Standard-form shortcuts

From Ax + By = C with B ≠ 0:

• Slope m = −A/B
• y-intercept = C/B, point (0, C/B)
• x-intercept = C/A if A ≠ 0, point (C/A, 0)

Worked example. 3x + 4y = 24. Slope = −3/4. y-intercept 24/4 = 6 → (0, 6). x-intercept 24/3 = 8 → (8, 0). Slope-intercept form: y = −(3/4)x + 6.

To convert point-slope into slope-intercept, distribute and add y₁. From slope 2 through (−3, 1): y − 1 = 2(x + 3) → y − 1 = 2x + 6 → y = 2x + 7. The y-intercept is 7, not 1. The 1 was only the known point's y-coordinate — and 1 will sit in the answer list if the question asked for b.

Finding a line from two points is a two-step habit:

1. Compute m from the two points.
2. Plug one of the points into point-slope (or into y = mx + b to solve for b).

Computing m and then stopping is how "2" becomes a trap when the stem asked for an intercept or a constant term.

Parallel and Perpendicular

• Parallel non-vertical lines: m₁ = m₂ and the y-intercepts differ. They never meet. The system of their equations has no solution.
• Same line (coincident): m₁ = m₂ and b₁ = b₂. Infinitely many solutions.
• Perpendicular non-vertical lines: slopes are negative reciprocals. m₁ · m₂ = −1, or m₂ = −1/m₁.
• Every horizontal line is perpendicular to every vertical line. That pair is the exception to "multiply the slopes to get −1," because one slope is undefined.

Negative reciprocal means two moves: flip the fraction and change the sign.

Worked example. Line L1 through (−3, 1) and (1, 9) has slope 2. Line L2 is perpendicular to L1 and passes through (6, 5). Then m₂ = −1/2. Point-slope: y − 5 = (−1/2)(x − 6) → y − 5 = −x/2 + 3 → y = −x/2 + 8. Set y = 0 for the x-intercept: 0 = −x/2 + 8, so x = 16. The value 8 is the y-intercept — the usual partial-calculation trap when the question asked for the x-intercept.

Perpendicular bisector (still coordinate algebra)

The perpendicular bisector of segment AB is the line through the midpoint of AB with slope −1/m_AB. Compute M from Chapter 6.1, flip the slope, write point-slope. That is midpoint plus negative reciprocal, not a Euclidean construction.

Worked example. Segment from (0, 0) to (6, 8). Midpoint (3, 4). Slope of the segment is 8/6 = 4/3, so the perpendicular slope is −3/4. Equation: y − 4 = (−3/4)(x − 3).

Intercepts and Intercept Triangles

• x-intercept: set y = 0 and solve for x. Point (x, 0).
• y-intercept: set x = 0 and solve for y. Point (0, y).

A line that is not horizontal, not vertical, and not through the origin cuts the axes at two points (a, 0) and (0, b). Those two intercepts and the origin form a right triangle with legs |a| and |b|. Its area is (1/2)|a · b|. That intercept triangle is coordinate algebra: two axis-parallel legs.

Worked example. The line 2x + 5y = 20 has intercepts (10, 0) and (0, 4). The triangle with the origin has area (1/2)(10)(4) = 20.

A second official-style setup is a triangle bounded by two lines and an axis.

Worked example. Region bounded by y = 2x, y = −(1/2)x + 5, and the y-axis x = 0.

• y = 2x meets x = 0 at (0, 0).
• y = −(1/2)x + 5 meets x = 0 at (0, 5).
• The two slanted lines: 2x = −x/2 + 5 → (5/2)x = 5 → x = 2, y = 4. Third vertex (2, 4).

Base along the y-axis has length 5. Height is the horizontal distance from (2, 4) to x = 0, which is 2. Area = (1/2)(5)(2) = 5. Forgetting the 1/2 produces 10; using the slanted segment as a "base" without an altitude from coordinates is how Euclidean instincts waste time on an algebra item.

Which quadrants does y = mx + b hit?

Any non-horizontal, non-vertical line that misses the origin passes through exactly three quadrants and misses one.

Memory: positive slope and positive y-intercept starts on the positive y-axis (between I and II) and rises to the right into I, so the missed quadrant is the bottom-right, IV. Negative slope from a positive y-intercept falls to the right into IV and never reaches III.

Worked example. kx + 3y = 12. Slope-intercept: y = (−k/3)x + 4. Here b = 4 > 0. The line hits I, II, and IV but not III precisely when the slope is negative: −k/3 < 0 → k > 0. If k < 0 the slope is positive and the line does enter III. If k = 0 the line is the horizontal y = 4 and stays in I and II only.

Intersections

Two lines with different slopes meet in exactly one point: solve the system. Equal slopes and different intercepts: no intersection (parallel). Equal slopes and equal intercepts: the same line.

Substitution is usually faster than elimination on Quant because one equation is often already solved for y. Set mx + b equal to the other expression, solve for x, then back-substitute for y. The solution pair is the intersection point on the plane — the same (x, y) you would plot.

Worked example. y = 3x − 1 and y = −x + 7. Then 3x − 1 = −x + 7 → 4x = 8 → x = 2, y = 5. Intersection (2, 5). Different slopes (3 and −1) guaranteed a unique meeting point; 3 · (−1) = −3, not −1, so the lines are not perpendicular.