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`        The equations of two sides of a square whose area is 25 square units are 3x-4y=0 and 4x+3y=0. The equations of the other two sides of the square are?`
one year ago

```							Dear student Since this is a square, the two other sides must be parallel to the first two sides. So they are: (3x−4y = a) and (4x+3y = b) We just need to find values for a and b. 3x−4y = 0 and 4x+3y = 0 intersect at point (0,0)Since square has area 25, side length = 5 So distance from (0,0) to lines (3x−4y = a) and (4x+3y = b) must = 5 Distance from (0,0) to line (3x−4y = a) = 5 |3(0) − 4(0) − a| / √(3²+(−4)²) = 5 |−a| / 5 = 5 |−a| = 25 a = ± 25 Distance from (0,0) to line (4x+3y = b) = 5 |4(0) + 3(0) − b| / √(4²+3²) = 5 |−b| / 5 = 5 |−b| = 25 b = ± 25 The other 2 lines are 3x−4y = ± 25 and 4x+3y = ± 25 There are 4 possible ways to choose the equations for the two other sides: 4x+3y = 25 and 3x−4y = 25 4x+3y = 25 and 3x−4y = −25 4x+3y = −25 and 3x−4y = 25 4x+3y = −25 and 3x−4y = −25  RegardsArun (askIITians forum expert)
```
one year ago
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