Question

in the diagram, points e, a, and b are collinear and the areas of square abcd and right ead are equal. what are the coordinates of a? e = (-10, 8) a = (?) b = (-7, 2) a. (-7.5, 3) b. (-8, 4) c. (-9, 6) d. (-8.5, 5) e. (-6, 5)

Explanation:

Step1: Let the coordinates of \(A\) be \((x,y)\)

Let \(A=(x,y)\), \(E = (- 10,8)\) and \(B=(-7,2)\)

Step2: Calculate the distance between two - points

The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). The length of \(AB=\sqrt{(x + 7)^2+(y - 2)^2}\), and the length of \(AE=\sqrt{(x + 10)^2+(y - 8)^2}\).
The area of right - triangle \(EAD\) is \(S_{EAD}=\frac{1}{2}\times AD\times AE\), and the area of square \(ABCD\) is \(S_{ABCD}=AD^2\). Since \(S_{EAD}=S_{ABCD}\), we have \(\frac{1}{2}\times AD\times AE=AD^2\), so \(AE = 2AD\).
Also, since \(\angle EAD = 90^{\circ}\), the slope of \(EA\) times the slope of \(AD\) is \(- 1\). The slope of the line passing through two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(m=\frac{y_2 - y_1}{x_2 - x_1}\). The slope of \(EA\) is \(m_{EA}=\frac{y - 8}{x + 10}\), and the slope of \(AB\) (since \(AD\perp AB\)) is \(m_{AB}=\frac{y - 2}{x + 7}\).
Another way is to use the vector method. \(\overrightarrow{EA}=(x + 10,y - 8)\) and \(\overrightarrow{AB}=(x + 7,y - 2)\). Since \(\angle EAB = 180^{\circ}\) (collinear points \(E\), \(A\), \(B\)) and \(\angle EAD=90^{\circ}\), we can also use the property of right - triangle and square area.
We can also check by substituting the options into the conditions:
For option B: If \(A=(-8,4)\)
The length of \(AB=\sqrt{(-8 + 7)^2+(4 - 2)^2}=\sqrt{1 + 4}=\sqrt{5}\)
The length of \(AE=\sqrt{(-8 + 10)^2+(4 - 8)^2}=\sqrt{4 + 16}=\sqrt{20}=2\sqrt{5}\)
The slope of \(EA=\frac{4 - 8}{-8+10}=\frac{-4}{2}=-2\)
The slope of \(AB=\frac{4 - 2}{-8 + 7}=\frac{2}{-1}=-2\) (Since \(E\), \(A\), \(B\) are collinear)
The area of right - triangle \(EAD\): Let \(AD = AB=\sqrt{5}\), \(AE = 2\sqrt{5}\), \(S_{EAD}=\frac{1}{2}\times AD\times AE=\frac{1}{2}\times\sqrt{5}\times2\sqrt{5}=5\)
The area of square \(ABCD\): \(S_{ABCD}=AB^2 = 5\)

Answer:

B. \((-8,4)\)