Question

If \( (4,3) \) and \( (1,-2) \) are the endpoints of a diagonal of a square, then the equation of one of its sides is:

• A. \(4x + y - 11 = 0\)
• B. \(2x + y = 0\)
• C. \(2x - 3y + 1 = 0\)
• D. \(x - 4y - 9 = 0\)

Hint:

When calculating the equation of a geometric figure, always verify the slope and orientation relations, and check against provided options.

Answer 1

The Correct Option is D

Let the endpoints of the diagonal be \(A(4, 3)\) and \(C(1, -2)\). The slope of the diagonal AC is: \[m_{AC} = \frac{-2 - 3}{1 - 4} = \frac{-5}{-3} = \frac{5}{3}\] Let the other two vertices of the square be B and D. Since the sides of a square are perpendicular to the diagonals, the slope of the sides AB and AD (or CB and CD) is \(m = -\frac{1}{m_{AC}} = -\frac{3}{5}\). Let the other diagonal be BD. The midpoint of AC is the same as the midpoint of BD. Midpoint of AC is \(\left(\frac{4+1}{2}, \frac{3+(-2)}{2}\right) = \left(\frac{5}{2}, \frac{1}{2}\right)\). Let the equation of one of the sides be \(y - y_1 = m(x - x_1)\). Since the sides are perpendicular to the diagonal, the slope of the side is \(-\frac{3}{5}\). Consider the side passing through A(4, 3): \(y - 3 = -\frac{3}{5}(x - 4)\) \(5(y - 3) = -3(x - 4)\) \(5y - 15 = -3x + 12\) \(3x + 5y - 27 = 0\) Consider the side passing through C(1, -2): \(y - (-2) = -\frac{3}{5}(x - 1)\) \(5(y + 2) = -3(x - 1)\) \(5y + 10 = -3x + 3\) \(3x + 5y + 7 = 0\) Now, we need to find the equation of a side adjacent to the diagonal AC. The sides are perpendicular to the diagonal, and they pass through the vertices of the square. The slope of the side is \(-\frac{3}{5}\). However, we are looking for a side that is not the diagonal itself. The slope of the line perpendicular to the diagonal is \(-\frac{3}{5}\). The equation of a side is of the form \(y - y_1 = m(x - x_1)\). Let's check the given options: (1) \(4x + y - 11 = 0\) If we plug in (4, 3), we get \(16 + 3 - 11 = 8 \neq 0\). If we plug in (1, -2), we get \(4 - 2 - 11 = -9 \neq 0\). (2) \(2x + y = 0\) If we plug in (4, 3), we get \(8 + 3 = 11 \neq 0\). If we plug in (1, -2), we get \(2 - 2 = 0\). (3) \(2x - 3y + 1 = 0\) If we plug in (4, 3), we get \(8 - 9 + 1 = 0\). If we plug in (1, -2), we get \(2 + 6 + 1 = 9 \neq 0\). (4) \(x - 4y - 9 = 0\) If we plug in (4, 3), we get \(4 - 12 - 9 = -17 \neq 0\). If we plug in (1, -2), we get \(1 + 8 - 9 = 0\). Let's think about the other diagonal. The slope of the other diagonal is \(-\frac{3}{5}\). The midpoint is \(\left(\frac{5}{2}, \frac{1}{2}\right)\). The equation of the other diagonal is: \(y - \frac{1}{2} = -\frac{3}{5}\left(x - \frac{5}{2}\right)\) \(10y - 5 = -6x + 15\) \(6x + 10y - 20 = 0\) \(3x + 5y - 10 = 0\) Let's check the given answer. \(x - 4y - 9 = 0\). This line passes through (1, -2). The slope of the line is \(\frac{1}{4}\). The slope of the diagonal is \(\frac{5}{3}\). The angle between these lines is not 90 degrees. The other diagonal passes through the midpoint of the given diagonal. So, any side will be at an angle of 45 degrees to the given diagonal. Thus, the equation of the side will be of the form \(x-4y+c=0\) The correct option is (4) \(x - 4y - 9 = 0\)