Coordinates, Straight Lines and Feasible Regions

A graph is another language for the relationships we have been calculating. One position records two values at once, allowing us to see patterns that are harder to notice in a list. Start every drawing by naming the quantities on the axes. “Quantity, in kilograms” and “cost, in pounds” are more useful labels than x and y alone. We conventionally put x horizontally and y vertically, but economic graphs may use other variable names or particular axis conventions. Read the labels rather than assuming. Every numerical graph in this lesson is an invented teaching model, not evidence about an actual market.

“Ordered” matters because reversing the entries usually places a point somewhere else. Three across and seven up is different from seven across and three up. If negative values are allowed, move left for negative x and down for negative y. Choose scales that fit the relevant values comfortably on the page. For example, each horizontal tick could represent one unit while each vertical tick represents five pounds. That is perfectly valid when labelled clearly. However, switching halfway along one axis from increments of five to increments of ten would distort an ordinary linear graph unless a special break or different scale were explicitly explained.

A small value list is a bridge between algebra and a graph. Substitute each horizontal value into the formula and preserve the matching pair. Three points are useful because the third can reveal an arithmetic or plotting error, even though two distinct points determine a straight line. If the variable can take every intermediate value, connect the points with a straight line over the permitted range. If only whole-number quantities are possible, individual admissible points have the literal meaning; the connecting line is a useful model guide. Do not extend a line beyond the domain without checking whether the formula still applies there.

The subscripts simply label the first and second points. They do not mean multiplication or powers. We calculate the vertical difference and divide it by the horizontal difference. In this example, moving three units right raises the vertical value by six, so each horizontal unit corresponds to two vertical units. You may subtract the first point from the second or reverse both subtractions; either gives the same slope. Reversing only one subtraction gives the wrong sign. Keep units: if y is cost in pounds and x is quantity in units, the slope is pounds per unit, not a unit-free number.

The two intercepts require different substitutions. On the vertical axis the horizontal coordinate is zero, so set x to zero. On the horizontal axis the vertical coordinate is zero, so set y to zero. In our example the crossings are the points zero comma ten and five comma zero. The slope is minus two, meaning a one-unit horizontal increase is associated with a two-unit vertical decrease. An algebraic intercept may lie outside the meaningful domain. A cost line with a positive fixed charge and positive slope, for instance, might have a negative horizontal intercept that does not describe a feasible production quantity.

Do not confuse a negative slope with a negative value. A falling line can remain entirely above the horizontal axis. Similarly, a rising line can include negative vertical values. Slope describes how values change, not where the line sits. A horizontal line has no vertical change, so its slope is zero divided by a non-zero horizontal change. A vertical line has the opposite issue: the horizontal denominator is zero, so its slope is undefined rather than zero. Finally, stretching the drawing vertically can make the same relationship look steeper. Numerical slopes and stated units protect against misleading visual comparisons.

Consider a cost relationship with a fixed charge and a constant per-unit charge. Ordering more units under the same charges moves us along the existing cost line. Increasing the fixed charge changes the cost associated with every quantity, so the entire line shifts upwards. Increasing the per-unit charge instead changes how quickly cost rises with quantity; the line becomes steeper when the axis scales stay fixed. This distinction is important because an observed point moving upwards does not by itself tell us which mechanism occurred. We need to know whether quantity changed, the underlying terms changed, or both changed at once.

Draw quantity in kilograms horizontally and total cost in pounds vertically. A horizontal range from zero to six and a vertical range from zero to thirty will include all three points. Mark equal steps on each axis, then locate the calculated pairs. The zero-quantity point represents the fixed charge built into this illustrative formula; whether such a charge would really be payable with no delivery depends on the actual service terms. Here it is part of the stated model. Because kilograms are divisible, the relationship includes intermediate quantities. Join the points with a straight line over the non-negative quantity range being considered.

The slope calculation recovers the coefficient of q, which is a useful check that our plotted values are consistent with the formula. Four additional kilograms raise cost by twelve pounds, so each kilogram adds three pounds. The intercept of twelve pounds is the model's fixed charge, not the price per kilogram. Solving for a horizontal intercept produces minus four kilograms. This is a valid feature of the unrestricted mathematical line, but it is not a possible delivery mass. Draw the meaningful non-negative part as the model, and explain rather than physically interpreting any extension into negative quantities.

We are now graphing two goods rather than a quantity and its cost. Put beans on the horizontal axis and rice on the vertical axis. Each combination has a total cost: six pounds times bean kilograms plus four pounds times rice kilograms. The budget allows spending less than or exactly forty-eight pounds, which explains the less-than-or-equal sign. To draw the boundary, temporarily replace that sign with equality. Set y to zero: six x equals forty-eight. Divide both sides by six: x = 48 ÷ 6 = 8. Set x to zero: four y equals forty-eight. Divide both sides by four: y = 48 ÷ 4 = 12. Join the intercepts eight comma zero and zero comma twelve.

A feasible combination satisfies every stated restriction. The first test point spends thirty-two pounds and leaves sixteen pounds unused, so it lies inside the affordable region. The second spends fifty-two pounds and exceeds the budget by four pounds. The origin is also affordable because buying neither good costs zero in this model. Use a solid boundary because equality is included: combinations spending exactly forty-eight pounds are allowed. Rearranging the boundary gives four y equals forty-eight minus six x, then y equals twelve minus one point five x. Its slope means an extra kilogram of beans requires giving up one point five kilograms of rice when spending remains at the budget limit.

The additional restriction might represent a five-kilogram availability limit in this invented scenario. Draw the vertical line x equals five and retain only points to its left, including the line itself. Then keep only points that also satisfy the budget and non-negativity conditions. The word “and” is important: satisfying one constraint is insufficient. The four listed corners outline the intersection. Fractional coordinates are acceptable here because kilograms are divisible. With indivisible items, only permitted whole-number points within the region would describe actual choices. A feasible region shows what is allowed; it does not identify a best choice until an objective or preference is specified.

A graph can be arithmetically accurate while encouraging a mistaken impression. Starting the vertical axis near the smallest plotted value can make a modest difference look dramatic, so inspect the tick labels rather than judging height alone. Distinguish a theoretical relationship based on stated assumptions from a scatter of observed values. Observed quantities may move together because a third factor affects both, because influence runs in the opposite direction, or because of selection and chance. Interpolation uses the relationship within its supported range; extrapolation extends beyond that range. Neither a straight line nor a neat pattern guarantees that such an extension will remain valid.

Connect this routine to the earlier lessons. Units tell you what each axis means. Algebra supplies the points, intercepts and constraints. Ratios underlie slope, while percentage changes offer a different way to compare movements when the base is known. For an inequality with a strict symbol, use a dashed boundary to show that equality is excluded; for an inclusive symbol, use a solid boundary. A test point should not lie on the boundary, because it must distinguish the two sides. In the exercises, calculate from the stated relationships rather than judging a sketch by eye, and keep any causal conclusion within the evidence actually supplied.