Question

mbc = \frac{e - 0}{1 - 0} = \frac{e}{1} = e application of the slope formula

1. draw the vertical line segment ac construction
2. \angle abc is a right angle definition of perpendicular lines
3. \triangle abc is a right triangle definition of a right triangle
4. ba = \sqrt{1 + d^{2}}

bc = \sqrt{e^{2} + 1}
ca = \sqrt{(d - e)^{2}} = d - e application of the distance formula

1. (\sqrt{1 + d^{2}})^{2} + (\sqrt{e^{2} + 1})^{2} =? pythagorean theorem

(1 + d^{2}) + (e^{2} + 1) = d^{2} - 2de + e^{2}

1. 2 + d^{2} + e^{2} = d^{2} - 2de + e^{2} simplify

2 = -2de
-1 = de

1. -1 = m_{ab}m_{bc} substitution property of equality

which expression is missing from step 7?
○ a. a^{2} + b^{2}
○ b. -2de
○ c. (a + b)^{2}
○ d. (d - e)^{2}

Explanation:

Step1: Recall the Pythagorean theorem

In a right - triangle \( \triangle ABC\) with legs \(a\) and \(b\) and hypotenuse \(c\), the Pythagorean theorem states that \(a^{2}+b^{2}=c^{2}\). In \( \triangle ABC\), if \(BA=\sqrt{1 + d^{2}}\), \(BC=\sqrt{e^{2}+1}\) and \(CA=\sqrt{(d - e)^{2}}=d - e\) (assuming \(d\geq e\)), then by the Pythagorean theorem \((\sqrt{1 + d^{2}})^{2}+(\sqrt{e^{2}+1})^{2}=CA^{2}\).

Step2: Substitute the value of \(CA\)

Since \(CA=\sqrt{(d - e)^{2}}\), then \(CA^{2}=(d - e)^{2}\)

Answer:

D. \((d - e)^{2}\)