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Geometry Difficulty 2.9 Junior Find the answer

A point is equidistant from the coordinate axes if the vertical distance from the point to the xx-axis is equal to the horizontal distance from the point to the yy-axis. The point of intersection of the vertical line x=ax = a with the line with equation 3x+8y=243x + 8y = 24 is equidistant from the coordinate axes. What is the sum of all possible values of aa?

A number or a short expression. Spacing and $ signs are ignored.

Solution

If a>0a > 0, the distance from the vertical line with equation x=ax = a to the yy-axis is aa. If a<0a < 0, the distance from the vertical line with equation x=ax = a to the yy-axis is a-a. In each case, there are exactly two points on the vertical line with equation x=ax = a that are also a distance of aa or a-a (as appropriate) from the xx-axis: (a,a)(a, a) and (a,a)(a, -a). These points lie on the horizontal lines with equations y=ay = a and y=ay = -a, respectively. If the point (a,a)(a, a) lies on the line 3x+8y=243x + 8y = 24, then 3a+8a=243a + 8a = 24 or a=2411a = \frac{24}{11}. If the point (a,a)(a, -a) lies on the line 3x+8y=243x + 8y = 24, then 3a8a=243a - 8a = 24 or a=245a = -\frac{24}{5}. The sum of these values of aa is 2411+(245)=12026455=14455\frac{24}{11} + \left(-\frac{24}{5}\right) = \frac{120 - 264}{55} = -\frac{144}{55}.

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.