### Video Transcript

Which of the following represents the equation of a straight line in two-intercept form? Is it (A) π₯ over π plus π¦ over π equals one? (B) π₯ over π plus π¦ over π equals π. (C) π¦ equals ππ₯ plus π. (D) ππ₯ plus ππ¦ plus π equals zero. Or (E) ππ₯ plus ππ¦ equals one.

We recall that there are many ways of writing the equation of a straight line. For example, option (D) ππ₯ plus ππ¦ plus π equals zero is known as the general form of the equation of a straight line. As we are looking for two-intercept form, we can rule out this option. Option (C) is written in slopeβintercept form. This is also sometimes written as π¦ equals ππ₯ plus π, where π is the slope or gradient of the line and π, or π in this example, is the π¦-intercept. We can therefore also rule out this option.

Letβs consider the straight line that intercepts the π₯- and π¦-axis as shown. If this line intercepts the π₯-axis at π and the π¦-axis at π, we know that the points of intersection have coordinates π, zero and zero, π. We then define the two-intercept form as follows. The two-intercept form of the equation of the straight line that intercepts the π₯-axis at π, zero and π¦-axis at zero, π is π₯ over π plus π¦ over π equals one. The correct answer is option (A).

We can derive this answer as follows. We begin by recalling the slope or gradient of a line is equal to π¦ sub two minus π¦ sub one over π₯ sub two minus π₯ sub one, where π₯ sub one, π¦ sub one and π₯ sub two, π¦ sub two are two points that lie on the line. This is also sometimes referred to as the change in π¦ over the change in π₯ or the rise over the run. From our diagram, we see that π is equal to π minus zero over zero minus π. This simplifies to π over negative π, which can also be written as negative π over π.

Next, we recall the pointβslope form for the equation of a straight line. This states that π¦ minus π¦ sub one is equal to π multiplied by π₯ minus π₯ sub one. Substituting in the values of π, π₯ sub one, and π¦ sub one, we have π¦ minus zero is equal to negative π over π multiplied by π₯ minus π. Distributing the parentheses or expanding the brackets on the right-hand side, we have π¦ is equal to negative π over π π₯ plus π. We can then divide through by π, giving us π¦ over π is equal to negative π₯ over π plus one. Adding π₯ over π to both sides gives us the required equation π₯ over π plus π¦ over π is equal to one. This is the two-intercept form of the equation of a straight line which intercepts the π₯-axis at π, zero and the π¦-axis at zero, π.