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Question
If two opposite vertices of a square are (1, 2) and (5, 8), find the coordinates of its other two vertices and the equations of its sides.
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Solution
Slope of AC = \[\frac{8 - 2}{5 - 1} = \frac{3}{2}\]
The sides AB and AD pass through the point A(1,2) and make an angle of \[{45}^\circ\] with AC whose slope is \[\frac{3}{2}\].
Equations of AB and AD are given by \[y - 2 = \frac{\frac{3}{2} \pm \tan {45}^\circ}{1 \mp \frac{3}{2}\tan {45}^\circ}\left( x - 1 \right)\]
\[\Rightarrow y - 2 = \frac{3 \pm 2}{2 \mp 3}\left( x - 1 \right)\]
\[\Rightarrow y - 2 = - 5\left( x - 1 \right) \text { and } y - 2 = \frac{1}{5}\left( x - 1 \right)\]
\[ \Rightarrow 5x + y - 7 = 0 \text { and } x - 5y + 9 = 0\]
Thus, the equations of AB and AD are \[5x + y - 7 = 0 \text { and } x - 5y + 9 = 0\] respectively.
Since BC is parallel to AD, the equation of BC is \[x - 5y + \lambda = 0\].
This line passes through C (5,8).
\[5 - 40 + \lambda = 0 \Rightarrow \lambda = 35\]
So, the equation of BC is \[x - 5y + 35 = 0\].
Since CD is parallel to AB, the equation of CD is \[5x + y + \lambda = 0\].
This line passes through C (5, 8).
\[25 + 8 + \lambda = 0 \Rightarrow \lambda = - 33\]
So, the equation of CD is \[5x + y - 33 = 0\].
Solving equation of AB and BC, we get B as (0, 7).
Solving equation of AD and CD, we get D as (6, 3).
Hence, the other two vertices are (0, 7) and (6, 3).
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| (a) The coordinates of the points P and Q on the line x + 5y = 13 which are at a distance of 2 units from the line 12x – 5y + 26 = 0 are |
(i) (3, 1), (–7, 11) |
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(ii) `(- 1/3, 11/3), (4/3, 7/3)` |
| (c) The coordinates of the point on the line joining A (–2, 5) and B (3, 1) such that AP = PQ = QB are |
(iii) `(1, 12/5), (-3, 16/5)` |
