The line \( y=a x+b \) passes through \( (1,-2) \) and is perpendicular to the line \( y-3 x=4 \). Find the value of \( a \) and of \( b \).

To find the values of \( a \) and \( b \), we first need to determine the slope of the line given by the equation \( y - 3x = 4 \). We can rewrite this in slope-intercept form: \[ y = 3x + 4 \] From this, we see that the slope \( m_1 \) of the line is \( 3 \). For the line \( y = ax + b \) to be perpendicular to this line, its slope \( a \) must satisfy the condition that the product of the slopes of the two lines is \( -1 \): \[ a \cdot 3 = -1 \implies a = -\frac{1}{3} \] Now we have the slope \( a = -\frac{1}{3} \). Next, we need to find \( b \) using the point \( (1, -2) \) which lies on the line \( y = ax + b \). Substituting \( x = 1 \) and \( y = -2 \) into the equation \( y = ax + b \): \[ -2 = -\frac{1}{3}(1) + b \] Solving for \( b \): \[ -2 = -\frac{1}{3} + b \] \[ b = -2 + \frac{1}{3} \] \[ b = -\frac{6}{3} + \frac{1}{3} = -\frac{5}{3} \] Thus, the values of \( a \) and \( b \) are: \[ a = -\frac{1}{3}, \quad b = -\frac{5}{3} \]