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Question
If one diagonal of a square is along the line 8x – 15y = 0 and one of its vertex is at (1, 2), then find the equation of sides of the square passing through this vertex.
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Solution
Let ABCD be the given square and the coordinates of the vertex D be (1, 2).
We are required to find the equations of its sides DC and AD.
Given that BD is along the line 8x – 15y = 0, so its slope is `8/15` (Why?).
The angles made by BD with sides AD and DC is 45° (Why?).
Let the slope of DC be m.
Then tan 45° = `(m - 8/15)/(1 + (8m)/15)` (Why?)
or 15 + 8m = 15m – 8
or 7m = 23, which gives m = `23/7`
Therefore, the equation of the side DC is given by
y – 2 = `23/7 (x - 1)` or 23x – 7y – 9 = 0.
Similarly, the equation of another side AD is given by
y – 2 = `(-7)/23 (x - 1)` or 7x + 23y – 53 = 0.
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