Given two lines Ax + 3y − 6 = 0 and 5x + 7y − k = 0, find the values of A and K if the lines are perpendicular and the second line passes through (4,3).

The general form of a linear equation is $Ax + By + C = 0$, and its slope $m$ is given by $m = -\frac{A}{B}$.

For the first line: $Ax + 3y - 6 = 0$
The slope $m_1 = -\frac{A}{3}$.

For the second line: $5x + 7y - k = 0$
The slope $m_2 = -\frac{5}{7}$.

If two lines are perpendicular, the product of their slopes is $-1$.

$$m_1 \cdot m_2 = -1$$

Substitute the slopes:

$$\left(-\frac{A}{3}\right) \cdot \left(-\frac{5}{7}\right) = -1$$ $$\frac{5A}{21} = -1$$

Multiply both sides by 21:

$$5A = -21$$

Divide by 5:

$$A = -\frac{21}{5}$$

Substitute $x=4$ and $y=3$ into the equation of the second line:

$$5(4) + 7(3) - k = 0$$ $$20 + 21 - k = 0$$ $$41 - k = 0$$

Add $k$ to both sides:

$$k = 41$$

The values are $A = -\frac{21}{5}$ and $K = 41$.

The values are $A = -\frac{21}{5}, K = 41$.