The points \(A(2,1)\), \(B(6,1)\) and \(C(6,5)\) are three vertices of a square \(ABCD\). Find the coordinates of \(D\) and write down the equation of the l...
Question 1 Report
The points \(A(2,1)\), \(B(6,1)\) and \(C(6,5)\) are three vertices of a square \(ABCD\).
Find the coordinates of \(D\) and write down the equation of the line of symmetry of the square that is parallel to the \(x\)-axis.
Answer Details
Finding \(D\). The given vertices are \(A(2,1)\), \(B(6,1)\) and \(C(6,5)\). Going \(A\) to \(B\) moves 4 units right, and \(B\) to \(C\) moves 4 units up, confirming a square of side 4. In the square \(ABCD\) the vertices are taken in order, so \(D\) must complete the circuit: from \(C\) move 4 units left (the reverse of \(A\) to \(B\)) to reach \(D\), or equivalently from \(A\) move 4 units up.
\[ D = (6-4,\; 5) = (2,5) \] [B1]The horizontal line of symmetry. A line parallel to the \(x\)-axis is horizontal, so its equation has the form \(y=\) a number. The square has its horizontal sides at \(y=1\) (through \(A\) and \(B\)) and \(y=5\) (through \(C\) and \(D\)), so the mirror line lies midway between them:
\[ y = \frac{1+5}{2} = 3 \]Equation: \( y = 3 \) [B1] (or any equivalent form)
The check that the square really is a square matters: only then does the midway horizontal line pass through the centre and act as a mirror. A quick sketch on axes prevents the common slip of writing \(x=3\) for a line parallel to the \(x\)-axis.
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