Question

question 1 (multiple choice worth 1 points) (64 - 01 rm c) δabc is located at a(2, 3), b(4, 4), and c(6, 3). zackery says that δabc is an isosceles triangle, while verna says that it is a right triangle. who is correct? zackery, because (overline{bc}congoverline{ac}) zackery, because (overline{ab}congoverline{bc}) verna, because (overline{bc}perpoverline{ac}) verna, because (overline{ab}perpoverline{bc})

Explanation:

Step1: Calculate the lengths of the sides

Use the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)
For \(AB\): \(x_1 = 2,y_1 = 3,x_2 = 4,y_2 = 4\)
\(AB=\sqrt{(4 - 2)^2+(4 - 3)^2}=\sqrt{4 + 1}=\sqrt{5}\)
For \(BC\): \(x_1 = 4,y_1 = 4,x_2 = 6,y_2 = 3\)
\(BC=\sqrt{(6 - 4)^2+(3 - 4)^2}=\sqrt{4+1}=\sqrt{5}\)
For \(AC\): \(x_1 = 2,y_1 = 3,x_2 = 6,y_2 = 3\)
\(AC=\sqrt{(6 - 2)^2+(3 - 3)^2}=\sqrt{16+0}=4\)

Step2: Check for isosceles and right - triangle properties

Since \(AB = BC=\sqrt{5}\), the triangle has two equal sides, so it is isosceles.
Check the slopes for right - triangle property.
Slope of \(AB\): \(m_{AB}=\frac{4 - 3}{4 - 2}=\frac{1}{2}\)
Slope of \(BC\): \(m_{BC}=\frac{3 - 4}{6 - 4}=-\frac{1}{2}\)
Slope of \(AC\): \(m_{AC}=\frac{3 - 3}{6 - 2}=0\)
The product of slopes of \(AB\) and \(BC\) is \(\frac{1}{2}\times(-\frac{1}{2})=-\frac{1}{4}
eq - 1\) (not perpendicular)
The product of slopes of \(BC\) and \(AC\) is \((-\frac{1}{2})\times0 = 0\) (not perpendicular)
The product of slopes of \(AB\) and \(AC\) is \(\frac{1}{2}\times0 = 0\) (not perpendicular)

Answer:

Zackery, because \(\overline{AB}\cong\overline{BC}\)