Question

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 coordinates of points

Point \(A(1,2)\), \(B(4,2)\), \(C(4, - 1)\). 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 at \(B\)

Since \(AB\) is horizontal and \(BC\) is vertical, the angle between \(AB\) and \(BC\) is \(90^{\circ}\). Also, we can calculate the lengths: \(AB=\vert4 - 1\vert=3\), \(BC=\vert2-(-1)\vert = 3\)? Wait, no, \(BC=\vert2-(-1)\vert=3\)? Wait, \(y\) - coordinate of \(B\) is \(2\), \(y\) - coordinate of \(C\) is \(- 1\), so \(BC=\vert2-(-1)\vert = 3\), and \(AB=\vert4 - 1\vert=3\)? Wait, no, \(AB\) is horizontal, length is difference in \(x\) - coordinates: \(AB = 4 - 1=3\), \(BC\) is vertical, length is difference in \(y\) - coordinates: \(BC=2-(-1)=3\). And since \(\angle B = 90^{\circ}\), the triangle \(ABC\) is a right - triangle.

Step 1: Recall the formula for the area of a right - triangle

The formula for the area of a right - triangle is \(A=\frac{1}{2}\times\text{base}\times\text{height}\).

Step 2: Identify the base and height

In right - triangle \(ABC\), the base \(AB\) (horizontal side) has length \(AB=\vert4 - 1\vert = 3\) (since the \(y\) - coordinates of \(A\) and \(B\) are the same, we calculate the difference in \(x\) - coordinates). The height \(BC\) (vertical side) has length \(BC=\vert2-(-1)\vert=3\) (since the \(x\) - coordinates of \(B\) and \(C\) are the same, we calculate the difference in \(y\) - coordinates).

Step 3: Calculate the area

Using the formula \(A=\frac{1}{2}\times AB\times BC\), substitute \(AB = 3\) and \(BC = 3\) into the formula: \(A=\frac{1}{2}\times3\times3=\frac{9}{2}=4.5\).

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. Also, the vector \(\overrightarrow{AB}\) should be equal to \(\overrightarrow{DC}\) and \(\overrightarrow{AD}\) should be equal to \(\overrightarrow{BC}\).
We know that \(A(1,2)\), \(B(4,2)\), \(C(4,-1)\). Let \(D(x,y)\).
Since \(\overrightarrow{AB}=(4 - 1,2 - 2)=(3,0)\) and \(\overrightarrow{DC}=(4 - x,-1 - y)\), for \(\overrightarrow{AB}=\overrightarrow{DC}\), we have \(4 - x = 3\) and \(-1 - y=0\). From \(4 - x = 3\), we get \(x = 1\); from \(-1 - y = 0\), we get \(y=-1\).
Or we can use the fact that in a rectangle, the diagonals bisect each other. The mid - point of \(AC\) is the same as the mid - point of \(BD\). The mid - point of \(AC\) with \(A(1,2)\) and \(C(4,-1)\) is \((\frac{1 + 4}{2},\frac{2+( - 1)}{2})=(\frac{5}{2},\frac{1}{2})\). Let \(D(x,y)\) and \(B(4,2)\), the mid - point of \(BD\) is \((\frac{4 + x}{2},\frac{2 + y}{2})\). So \(\frac{4 + x}{2}=\frac{5}{2}\) and \(\frac{2 + y}{2}=\frac{1}{2}\). Solving \(\frac{4 + x}{2}=\frac{5}{2}\), we multiply both sides by \(2\): \(4+x = 5\), so \(x = 1\). Solving \(\frac{2 + y}{2}=\frac{1}{2}\), we multiply both sides by \(2\): \(2 + y=1\), so \(y=-1\).

Step 2: Verify the sides

For \(ABCD\) with \(A(1,2)\), \(B(4,2)\), \(C(4,-1)\), \(D(1,-1)\):

• \(AB\): length is \(4 - 1 = 3\), slope is \(\frac{2 - 2}{4 - 1}=0\) (horizontal line).
• \(DC\): length is \(4 - 1 = 3\), slope is \(\frac{-1-(-1)}{4 - 1}=0\) (horizontal line), so \(AB\parallel DC\) and \(AB = DC\).
• \(AD\): length is \(\vert2-(-1)\vert = 3\), slope is \(\frac{-1 - 2}{1 - 1}\) (undefined, vertical line).
• \(BC\): length is \(\vert2-(-1)\vert = 3\), slope is \(\frac{-1 - 2}{4 - 4}\) (undefined, vertical line), so \(AD\parallel BC\) and \(AD = BC\). Also, \(AB\perp AD\) (since \(AB\) is horizontal and \(AD\) is vertical).

Answer:

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

Sub - Question 2