Question

in the coordinate plane, points a, b, and c have coordinates (1,2), (4,2), and (4, -1) respectively.

1. plot points a, b, and c on a coordinate grid, then connect them to form a figure. what is the shape of this figure?
2. calculate the area of the figure formed by points a, b, and c.
3. find the coordinates of point d such that quadrilateral abcd is a rectangle. plot point d and verify the rectangle by checking the properties of its sides.

Explanation:

Sub - question 1

Step 1: Analyze the coordinates

For point \(A(1,2)\), \(B(4,2)\) and \(C(4, - 1)\). The \(y\) - coordinate of \(A\) and \(B\) is the same (\(y = 2\)), so the line segment \(AB\) is horizontal. The \(x\) - coordinate of \(B\) and \(C\) is the same (\(x=4\)), so the line segment \(BC\) is vertical.

Step 2: Determine the angle

Since \(AB\) is horizontal and \(BC\) is vertical, the angle between \(AB\) and \(BC\) is \(90^{\circ}\). And we have three points, so the figure formed by \(A\), \(B\), \(C\) is a right - triangle.

Step 1: Calculate the length of \(AB\)

The length of a horizontal line segment with endpoints \((x_1,y)\) and \((x_2,y)\) is given by \(d=\vert x_2 - x_1\vert\). For \(A(1,2)\) and \(B(4,2)\), \(AB=\vert4 - 1\vert=3\).

Step 2: Calculate the length of \(BC\)

The length of a vertical line segment with endpoints \((x,y_1)\) and \((x,y_2)\) is given by \(d = \vert y_2 - y_1\vert\). For \(B(4,2)\) and \(C(4,-1)\), \(BC=\vert-1 - 2\vert=\vert-3\vert = 3\).

Step 3: Calculate the area of the right - triangle

The area of a right - triangle is given by \(A=\frac{1}{2}\times\text{base}\times\text{height}\). Here, the base \(AB = 3\) and the height \(BC=3\). So \(A=\frac{1}{2}\times3\times3=\frac{9}{2}=4.5\).

Step 1: Recall the properties of a rectangle

In a rectangle \(ABCD\), the opposite sides are equal and parallel, and the adjacent sides are perpendicular. Since \(ABCD\) is a rectangle, \(\overrightarrow{AB}=\overrightarrow{DC}\) and \(\overrightarrow{AD}=\overrightarrow{BC}\).
We know that \(\overrightarrow{AB}=(4 - 1,2 - 2)=(3,0)\) and \(\overrightarrow{BC}=(4 - 4,-1 - 2)=(0,-3)\).
If \(A(1,2)\), \(B(4,2)\), \(C(4,-1)\), let \(D(x,y)\). Since \(\overrightarrow{AD}=\overrightarrow{BC}\), \((x - 1,y - 2)=(0,-3)\).

Step 2: Solve for the coordinates of \(D\)

From \(x - 1=0\), we get \(x = 1\). From \(y - 2=-3\), we get \(y=2-3=-1\). So the coordinates of \(D\) are \((1,-1)\).
To verify:

• The length of \(AB\): Using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), for \(A(1,2)\) and \(B(4,2)\), \(AB=\sqrt{(4 - 1)^2+(2 - 2)^2}=3\). For \(D(1,-1)\) and \(C(4,-1)\), \(DC=\sqrt{(4 - 1)^2+(-1+1)^2}=3\), so \(AB = DC\).
• The length of \(AD\): For \(A(1,2)\) and \(D(1,-1)\), \(AD=\sqrt{(1 - 1)^2+(-1 - 2)^2}=3\). For \(B(4,2)\) and \(C(4,-1)\), \(BC=\sqrt{(4 - 4)^2+(-1 - 2)^2}=3\), so \(AD = BC\).
• The slopes: The slope of \(AB\) is \(m_{AB}=\frac{2 - 2}{4 - 1}=0\) (horizontal line), the slope of \(BC\) is \(m_{BC}=\frac{-1 - 2}{4 - 4}\) (undefined, vertical line), the slope of \(CD\) is \(m_{CD}=\frac{-1+1}{1 - 4}=0\) (horizontal line), the slope of \(DA\) is \(m_{DA}=\frac{2 + 1}{1 - 1}\) (undefined, vertical line). So adjacent sides are perpendicular and opposite sides are parallel.

Answer:

The figure formed by points \(A\), \(B\), \(C\) is a right - triangle.

Sub - question 2