To determine if line [tex]\( \overline{QR} \)[/tex] is tangent to circle [tex]\( O \)[/tex] at point [tex]\( R \)[/tex], we need to check the relationship between the slopes of [tex]\( \overline{QR} \)[/tex] and [tex]\( \overline{OR} \)[/tex]. Specifically, if the slope of [tex]\( \overline{OR} \)[/tex] times the slope of [tex]\( \overline{QR} \)[/tex] equals [tex]\(-1\)[/tex], then they are perpendicular. A tangent to a circle at a point is perpendicular to the radius at that point.
Let's start by finding the slope of [tex]\( \overline{QR} \)[/tex]. The equation given is:
[tex]\[ 5y = -4x + 41 \][/tex]
We can rewrite this equation in the slope-intercept form [tex]\( y = mx + c \)[/tex], where [tex]\( m \)[/tex] is the slope:
[tex]\[ y = \left( -\frac{4}{5} \right) x + \frac{41}{5} \][/tex]
From this equation, we see that the slope [tex]\( m \)[/tex] of [tex]\( \overline{QR} \)[/tex] is:
[tex]\[ \text{slope of } \overline{QR} = -\frac{4}{5} \][/tex]
For [tex]\( \overline{QR} \)[/tex] to be tangent to circle [tex]\( O \)[/tex] at [tex]\( R \)[/tex], the slope of [tex]\( \overline{OR} \)[/tex] should be perpendicular to the slope of [tex]\( \overline{QR} \)[/tex]. This means their slopes should multiply to [tex]\(-1\)[/tex]:
[tex]\[ \text{slope of } \overline{OR} \times \left(-\frac{4}{5}\right) = -1 \][/tex]
Solving for the slope of [tex]\( \overline{OR} \)[/tex]:
[tex]\[ \text{slope of } \overline{OR} = \frac{5}{4} \][/tex]
Thus, the product of the slopes of [tex]\( \overline{QR} \)[/tex] and [tex]\( \overline{OR} \)[/tex] is:
[tex]\[ \left( \frac{5}{4} \right) \times \left( -\frac{4}{5} \right) = -1 \][/tex]
Since this condition is satisfied, it means that the line [tex]\( \overline{QR} \)[/tex] is tangent to the circle at point [tex]\( R \)[/tex].
Therefore, the correct answer is:
Yes, because the slope of [tex]\( \overline{OR} \)[/tex] times the slope of [tex]\( \overline{QR} \)[/tex] equals [tex]\(-1\)[/tex].
