Show that the points (1, 7), (4, 2), (–1, –1) and (– 4, 4) are the vertices of a square.
Step-by-Step Solution
Key Concept: In a square all four sides are equal and the two diagonals are equal. Using the distance formula we can compute the lengths of all possible line segments joining the given points. If we can arrange the points so that four equal sides and two equal diagonals are obtained, the quadrilateral is a square. Additionally, the product of slopes of adjacent sides must be –1 (perpendicular).
1. List the points:\
\(A(1,7),\; B(4,2),\; C(-1,-1),\; D(-4,4)\).
2. Compute distances between every pair using the distance formula \(AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\).
- \(AB = \sqrt{(4-1)^2+(2-7)^2}=\sqrt{3^2+(-5)^2}=\sqrt{9+25}=\sqrt{34}\)
- \(AD = \sqrt{(-4-1)^2+(4-7)^2}=\sqrt{(-5)^2+(-3)^2}=\sqrt{25+9}=\sqrt{34}\)
- \(BC = \sqrt{(-1-4)^2+(-1-2)^2}=\sqrt{(-5)^2+(-3)^2}=\sqrt{34}\)
- \(CD = \sqrt{(-4+1)^2+(4+1)^2}=\sqrt{(-3)^2+5^2}=\sqrt{34}\)
- \(AC = \sqrt{(-1-1)^2+(-1-7)^2}=\sqrt{(-2)^2+(-8)^2}=\sqrt{4+64}=\sqrt{68}\)
- \(BD = \sqrt{(-4-4)^2+(4-2)^2}=\sqrt{(-8)^2+2^2}=\sqrt{64+4}=\sqrt{68}\)
3. Identify the ordering: Since \(AB, BC, CD, DA\) are all \(\sqrt{34}\) and the two remaining distances \(AC\) and \(BD\) are \(\sqrt{68}\), the points can be taken in the order \(A\to B\to C\to D\to A\).
4. Check perpendicularity (optional but confirms square):
- Slope of \(AB\): \(m_{AB}=\frac{2-7}{4-1}=\frac{-5}{3}\).
- Slope of \(BC\): \(m_{BC}=\frac{-1-2}{-1-4}=\frac{-3}{-5}=\frac{3}{5}\).
- \(m_{AB}\times m_{BC}=\frac{-5}{3}\times\frac{3}{5}=-1\). Hence \(AB\perp BC\).
- Similarly \(m_{CD}=\frac{4+1}{-4+1}=\frac{5}{-3}= -\frac{5}{3}\) and \(m_{DA}=\frac{7-4}{1+4}=\frac{3}{5}\); their product is also \(-1\).
5. Conclusion: All four sides are equal (\(\sqrt{34}\)), the two diagonals are equal (\(\sqrt{68}\)), and adjacent sides are perpendicular. Therefore the quadrilateral formed by the given points is a square.
6. Answer: The points \((1,7), (4,2), (-1,-1), (-4,4)\) are indeed the vertices of a square.
Correct Answer: Yes, the four points are the vertices of a square (side = \(\sqrt{34}\), diagonal = \(\sqrt{68}\)).