To determine which line is perpendicular to a line that has a slope of [tex]\(-\frac{5}{6}\)[/tex], we need to figure out the slope of the perpendicular line.
For two lines to be perpendicular, the product of their slopes must be [tex]\(-1\)[/tex]. So, if the slope of the original line is [tex]\( m_1 = -\frac{5}{6} \)[/tex], the slope [tex]\( m_2 \)[/tex] of the perpendicular line can be calculated using the relationship:
[tex]\[ m_1 \times m_2 = -1 \][/tex]
Given the slope of the original line [tex]\( m_1 = -\frac{5}{6} \)[/tex], we substitute this into the equation:
[tex]\[ -\frac{5}{6} \times m_2 = -1 \][/tex]
To solve for [tex]\( m_2 \)[/tex], we rearrange the equation:
[tex]\[ m_2 = \frac{-1}{-\frac{5}{6}} \][/tex]
When we simplify [tex]\(\frac{-1}{-\frac{5}{6}}\)[/tex], we get:
[tex]\[ m_2 = \frac{6}{5} = 1.2 \][/tex]
So, the slope of the perpendicular line is [tex]\( 1.2 \)[/tex].
Thus, any line with a slope of [tex]\( 1.2 \)[/tex] will be perpendicular to a line with a slope of [tex]\( -\frac{5}{6} \)[/tex]. To identify which specific line (JK, LM, NO, PQ) corresponds to this slope, further information about the slopes of these lines would be required, which is not provided in the question. Therefore, we conclude that the perpendicular line to the given slope of [tex]\(-\frac{5}{6}\)[/tex] has a slope of [tex]\(1.2\)[/tex].
