To find the equation of line \( t \) that is parallel to line \( s \) and passes through the point \((-3, 3)\), we first need to identify the slope of line \( s \).
The equation of line \( s \) is given as: \[ y = -\frac{4}{3}x - 5 \]
From this equation, we can see that the slope (\( m \)) of line \( s \) is: \[ m_s = -\frac{4}{3} \]
Since line \( t \) is parallel to line \( s \), it will have the same slope: \[ m_t = -\frac{4}{3} \]
Next, we will use the point-slope formula to find the equation of line \( t \), which includes the point \((-3, 3)\). The point-slope form of the line's equation is: \[ y - y_1 = m(x - x_1) \] where \( (x_1, y_1) \) is the point the line passes through.
Substituting in \((-3, 3)\) for \( (x_1, y_1) \) and using \( m_t = -\frac{4}{3} \): \[ y - 3 = -\frac{4}{3}(x + 3) \]
Now, we simplify the equation:
First, distribute \( -\frac{4}{3} \): \[ y - 3 = -\frac{4}{3}x - 4 \]
Next, add 3 to both sides: \[ y = -\frac{4}{3}x - 4 + 3 \] \[ y = -\frac{4}{3}x - 1 \]
The equation of line \( t \) in slope-intercept form is: \[ \boxed{y = -\frac{4}{3}x - 1} \]
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