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Your friend says that [tex]\( y = 2x - 5 \)[/tex] is perpendicular to the line [tex]\( y = \frac{1}{2}x + 5 \)[/tex], which passes through [tex]\((2,6)\)[/tex]. She proves it with the following work:

[tex]\[
\begin{array}{l}
y = 2x - 5 \\
y = \frac{1}{2}x + b \\
6 = \frac{1}{2}(2) + b \\
6 = 1 + b \\
b = 5 \\
y = \frac{1}{2}x + 5
\end{array}
\][/tex]

Does her work make sense?


Sagot :

To determine whether your friend's work makes sense, we need to analyze her solution and check if the lines [tex]\( y = 2x - 5 \)[/tex] and [tex]\( y = \frac{1}{2}x + 5 \)[/tex] are indeed perpendicular.

1. Calculate the slopes of the lines:
- The slope of the first line [tex]\( y = 2x - 5 \)[/tex]:
The equation is already in the form [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] is the slope. Thus, the slope ([tex]\( m_1 \)[/tex]) is [tex]\( 2 \)[/tex].

- The slope of the second line [tex]\( y = \frac{1}{2}x + 5 \)[/tex]:
Similarly, this equation is in the form [tex]\( y = mx + b \)[/tex]. Therefore, the slope ([tex]\( m_2 \)[/tex]) is [tex]\( \frac{1}{2} \)[/tex].

2. Check the product of their slopes:
- For two lines to be perpendicular, the product of their slopes must be [tex]\( -1 \)[/tex].
- Calculate the product of the slopes:
[tex]\[ \text{Product of slopes} = m_1 \times m_2 = 2 \times \frac{1}{2} = 1 \][/tex]

3. Evaluate the result:
- The product of the slopes is [tex]\( 1 \)[/tex], not [tex]\( -1 \)[/tex]. Hence, the lines are not perpendicular.

4. Examine her algebraic steps briefly:
- While her algebraic manipulation has errors (particularly [tex]\( 6 = 1 + 3 \rightarrow 5 = 3 \)[/tex]), which are clearly incorrect, correcting these does not change the slopes of the lines.

Based on this analysis, her work does not make sense. The lines [tex]\( y = 2x - 5 \)[/tex] and [tex]\( y = \frac{1}{2}x + 5 \)[/tex] are not perpendicular because the product of their slopes is [tex]\( 1 \)[/tex], not [tex]\( -1 \)[/tex]. Therefore, the lines do not meet the perpendicularity condition.
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