Answer :
To determine which line is perpendicular to a given line with a slope of [tex]\(-\frac{5}{6}\)[/tex], we need to find the slope of the perpendicular line. Here is the step-by-step solution:
1. Identify the slope of the given line:
The slope of the given line is [tex]\(-\frac{5}{6}\)[/tex].
2. Calculate the slope of the perpendicular line:
The slope of any line that is perpendicular to another line is the negative reciprocal of the slope of the original line. The negative reciprocal of a fraction [tex]\(\frac{a}{b}\)[/tex] is [tex]\(-\frac{b}{a}\)[/tex]. For our given slope [tex]\(-\frac{5}{6}\)[/tex], the negative reciprocal is calculated as follows:
[tex]\[ \text{Perpendicular slope} = -\left(-\frac{6}{5}\right) = \frac{6}{5} \][/tex]
3. Convert the slope to decimal form if necessary:
The slope [tex]\(\frac{6}{5}\)[/tex] can be expressed in decimal form as 1.2.
Thus, the line that will be perpendicular to the line with a slope of [tex]\(-\frac{5}{6}\)[/tex] will have a slope of 1.2. Therefore, check the slopes of line JK, line LM, line NO, and line PQ to determine which one of them has a slope of 1.2.
1. Identify the slope of the given line:
The slope of the given line is [tex]\(-\frac{5}{6}\)[/tex].
2. Calculate the slope of the perpendicular line:
The slope of any line that is perpendicular to another line is the negative reciprocal of the slope of the original line. The negative reciprocal of a fraction [tex]\(\frac{a}{b}\)[/tex] is [tex]\(-\frac{b}{a}\)[/tex]. For our given slope [tex]\(-\frac{5}{6}\)[/tex], the negative reciprocal is calculated as follows:
[tex]\[ \text{Perpendicular slope} = -\left(-\frac{6}{5}\right) = \frac{6}{5} \][/tex]
3. Convert the slope to decimal form if necessary:
The slope [tex]\(\frac{6}{5}\)[/tex] can be expressed in decimal form as 1.2.
Thus, the line that will be perpendicular to the line with a slope of [tex]\(-\frac{5}{6}\)[/tex] will have a slope of 1.2. Therefore, check the slopes of line JK, line LM, line NO, and line PQ to determine which one of them has a slope of 1.2.
