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Question

The diagonal of a square lies along the line 8x − 15y = 0 and one vertex of the square is (1, 2). Find the equations to the sides of the square passing through this vertex.

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Solution

Let A (1, 2) be the vertex of square ABCD and BD be the diagonal that is along the line 8x − 15y = 0

Equation of the given line is,8x-15y=0-15y=-8xy=815xComparing this equation with y=mx+cwe get,m=815
So, the slope of BD will be 815.


Here, we have to find the equations of sides AB and AD.

We know that the equations of two lines passing through a point x1,y1 and making an angle α with the line whose slope is m.

y-y1=m±tanα1mtanαx-x1

Here,

m=815x1=1 y1=2 α=45

So, the equations of the required sides are

y-2=815+tan451-815tan45x-1 and y-2=815-tan451+815tan45x-1y-2=815+11-815x-1 and y-2=815-11+815x-123x-7y-9=0 and 7x+23y-53=0

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