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What is the distance between the parallel lines π₯ minus six π¦ plus 11 equals zero and π₯ minus six π¦ plus 22 equals zero?
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Firstly, letβs clarify what is meant by the distance between these parallel lines.
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There are of course a number of different distances between these two lines, depending on the two points that you choose to connect.
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When we refer to the distance between parallel lines, we mean the shortest distance between them, which is in fact the perpendicular distance.
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Letβs think about how to approach this.
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We have a formula for telling us how to calculate the perpendicular distance from a point π₯ one, π¦ one to a line with equation ππ₯ plus ππ¦ plus π equals zero.
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The distance is equal to the modulus of ππ₯ one plus ππ¦ one plus π divided by the square root of π squared plus π squared.
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We could apply this formula to calculating the distance between parallel lines if we can find the coordinates of a point that lies on one of the two lines.
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Letβs consider the first line π₯ minus six π¦ plus 11 is equal to zero.
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For simplicity, letβs find the coordinates of the point where this line meets the π¦-axis, its π¦-intercept.
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Adding six π¦ to both sides of this equation gives six π¦ is equal to π₯ plus 11.
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Dividing both sides of the equation by six gives π¦ is equal to π₯ over six plus 11 over six.
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Now if we can pair this with the slope intercept form of the equation of a straight line, we can see that the π¦-intercept of this line is 11 over six.
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As this is a point on the π¦-axis, its π₯-coordinate will be zero.
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And therefore, the coordinates of this point are zero, 11 over six.
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So what weβll do is use the coordinates of this point with the equation of the second line in order to calculate the distance between them.
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Letβs look back at this distance formula.
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The point with coordinates π₯ one, π¦ one is now the point with coordinates zero, 11 over six.
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By comparing the general form of the equation of the straight line with the straight line in this question, we can see that π is equal to one, π is equal to negative six, and π is equal to 22.
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Now letβs substitute the values of π, π, π, π₯ one, and π¦ one into the distance formula.
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We have that π is equal to the modulus or absolute value of one multiplied by zero plus negative six multiplied by 11 over six plus 22 all over the square root of one squared plus negative six squared.
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Now evaluating each of these parts and simplifying leads to the modulus of 11 over the square root of 37.
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Remember, the modulus of a number is just its absolute value.
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As 11 is already a positive number, the modulus of 11 is also 11.
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So we have that the distance between these two lines is 11 over root 37.
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Now this fraction currently has a surd in the denominator.
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So we need to rationalize it.
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And to do so, weβll multiply by root 37 over root 37.
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Doing so gives 11 root 37 in the numerator.
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And in the denominator, root 37 multiplied by root 37 is just 37.
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Therefore, the distance between the two parallel lines π₯ minus six π¦ plus 11 equals zero and π₯ minus six π¦ plus 22 equals zero is 11 root 37 over 37.