Overview
A line has both a direction and a location. Slope describes the change in height per change in horizontal position; an intercept or a point anchors the line. Confusing these roles leads to many plausible wrong answers. We will calculate a slope from two points, infer a perpendicular direction, place a parallel line through a new point, and recognize the special structure of a vertical line.
Keep the order consistent in rise over run
For A(−3, 4) and B(5, −2), the vertical change from A to B is −2 − 4 = −6, and the horizontal change is 5 − (−3) = 8. The slope is therefore −6/8 = −3/4. It is negative because moving right takes you downward.
You may subtract in the opposite order, but then both numerator and denominator must reverse. Changing only one reverses the sign incorrectly. The option −4/3 swaps rise and run; 3/4 loses the descending direction. A sketch gives a sign check without supplying exact measurement: the second point lies to the right and below the first. The coordinates, rather than the apparent steepness of a drawing, determine the exact slope. Before calculating, say which difference is vertical and which is horizontal so that the ratio keeps its intended meaning.
Find the original slope before taking a perpendicular one
For the line 2x + 5y = 10, isolate y to see 5y = −2x + 10 and therefore y = (−2/5)x + 2. Its slope is −2/5. A perpendicular nonvertical line has slope 5/2, the negative reciprocal, because the product of the two slopes is −1.
The answer −5/2 takes the reciprocal but omits the sign change. The answer 2/5 changes the sign but omits the reciprocal. Reading a coefficient directly from standard form would introduce a different mistake: the slope is not simply the coefficient of x before y is isolated. This sequence of two decisions—identify the original slope, then transform it—keeps the rule from becoming a guess based on the visible numbers. Vertical and horizontal lines require separate treatment because their slope values do not both fit the ordinary reciprocal calculation.
Use a point to locate a parallel line
A line parallel to y = 4x − 7 has slope 4. If it passes through (1, 2), write its equation as y = 4x + b and substitute the point: 2 = 4·1 + b. Hence b = −2, which is the requested y-intercept.
Parallel lines share direction, not location. Copying −7 would reproduce the original line, which does not pass through the new point. Copying 2 would confuse the point’s y-coordinate with the y-intercept; a y-intercept occurs specifically where x = 0. Once you find b, check that your line gives y = 2 when x = 1. You can also visualize the step back from the point to the y-axis: moving left one unit decreases y by four, taking the value from 2 to −2. That interpretation links the algebra to slope’s change-per-unit meaning.
Describe a vertical line by its fixed coordinate
Every point on the vertical line through (−4, 6) has x-coordinate −4, so its equation is x = −4. The y-coordinate can vary. The equation y = 6 instead describes a horizontal line, where height is fixed and x varies. A vertical line has no finite slope because its horizontal change is zero, so forcing it into y = mx + b would require division by zero.
For a new line question, first identify which features are supplied and which are requested. Two points determine a change ratio; a parallel or perpendicular relationship determines direction; one point then determines location. A vertical or horizontal description fixes a coordinate directly. During practice, state a quick geometric check after the algebra: whether the line rises or falls, whether it passes through the required point, and whether the shared feature is slope or intercept. These checks distinguish correct arithmetic from a correct interpretation of the line.
Apply the method to a new case
A line passes through (2, 3) and is perpendicular to the line y = 2x − 1. Where does it meet the y-axis? The original slope is 2, so the new slope is −1/2. Write y − 3 = −(x − 2)/2. At the y-axis, x = 0; this gives y = 4, so the intercept is (0, 4). Reporting −1/2 answers the slope question rather than the requested intersection. Using slope −2 negates without reciprocating and fails the perpendicular product check. Substitute (2, 3) into the new equation and verify that the two slopes multiply to −1. Those checks establish both the required point and the perpendicular relationship; neither check alone determines the intended line. Be explicit about whether the target is a slope, an equation or an ordered pair.