To solve for the slope of line [tex]\( r \)[/tex] that is perpendicular to line [tex]\( q \)[/tex], we'll first need to determine the slope of line [tex]\( q \)[/tex] itself.
The slope [tex]\( m \)[/tex] of a line passing through two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by the formula:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Given the points [tex]\((1, 10)\)[/tex] and [tex]\((9, 3)\)[/tex] for line [tex]\( q \)[/tex]:
- [tex]\( (x_1, y_1) = (1, 10) \)[/tex]
- [tex]\( (x_2, y_2) = (9, 3) \)[/tex]
Let's plug these values into the formula to get the slope of line [tex]\( q \)[/tex]:
[tex]\[ m_q = \frac{3 - 10}{9 - 1} \][/tex]
[tex]\[ m_q = \frac{-7}{8} \][/tex]
So, the slope of line [tex]\( q \)[/tex] is [tex]\( -\frac{7}{8} \)[/tex].
For a line to be perpendicular to another line, the product of their slopes must equal [tex]\(-1\)[/tex]. If [tex]\( m_q \)[/tex] is the slope of line [tex]\( q \)[/tex], then the slope of the perpendicular line [tex]\( r \)[/tex], denoted [tex]\( m_r \)[/tex], is given by:
[tex]\[ m_q \times m_r = -1 \][/tex]
Substituting [tex]\( m_q = -\frac{7}{8} \)[/tex]:
[tex]\[ \left(-\frac{7}{8}\right) \times m_r = -1 \][/tex]
To find [tex]\( m_r \)[/tex], divide both sides of the equation by [tex]\(-\frac{7}{8} \)[/tex]:
[tex]\[ m_r = \frac{-1}{-\frac{7}{8}} \][/tex]
When dividing by a fraction, multiply by its reciprocal:
[tex]\[ m_r = \frac{-1}{-\frac{7}{8}} = -1 \times -\frac{8}{7} = \frac{8}{7} \][/tex]
Therefore, the slope of line [tex]\( r \)[/tex] is:
[tex]\[ m_r = \frac{8}{7} \][/tex]
This is an improper fraction and is fully simplified. So the slope of the line perpendicular to line [tex]\( q \)[/tex] is [tex]\( \frac{8}{7} \)[/tex].
