WEBVTT
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Which of the following represents the equation of a straight line in two-intercept form?
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Is it (A) π₯ over π plus π¦ over π equals one?
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(B) π₯ over π plus π¦ over π equals π.
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(C) π¦ equals ππ₯ plus π.
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(D) ππ₯ plus ππ¦ plus π equals zero.
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Or (E) ππ₯ plus ππ¦ equals one.
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We recall that there are many ways of writing the equation of a straight line.
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For example, option (D) ππ₯ plus ππ¦ plus π equals zero is known as the general form of the equation of a straight line.
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As we are looking for two-intercept form, we can rule out this option.
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Option (C) is written in slopeβintercept form.
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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.
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We can therefore also rule out this option.
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Letβs consider the straight line that intercepts the π₯- and π¦-axis as shown.
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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.
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We then define the two-intercept form as follows.
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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.
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The correct answer is option (A).
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We can derive this answer as follows.
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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.
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This is also sometimes referred to as the change in π¦ over the change in π₯ or the rise over the run.
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From our diagram, we see that π is equal to π minus zero over zero minus π.
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This simplifies to π over negative π, which can also be written as negative π over π.
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Next, we recall the pointβslope form for the equation of a straight line.
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This states that π¦ minus π¦ sub one is equal to π multiplied by π₯ minus π₯ sub one.
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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.
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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.
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We can then divide through by π, giving us π¦ over π is equal to negative π₯ over π plus one.
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Adding π₯ over π to both sides gives us the required equation π₯ over π plus π¦ over π is equal to one.
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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.