Let $$Ax+By=C$$ and $$A'x+B'y=C'$$ represent two lines.

(a) If $$\displaystyle \frac{A}{A'} = \frac{B}{B'} \neq \frac{C}{C'}$$, then the lines are parallel.

Above properties provide us a way to write the equation of a line parallel or perpendicular to a given line that contains a given point not on the line. For example, suppose that we want the equation of the line pependicular to $$3x+4y=6$$ that contains the point (1, 2). The form $$4x-3y=k$$, where k is a constant, represents a family of lines perpendicular to $$3x+4y=6$$ because we have satisfied the condition $$AA' = - BB'$$. Therefore, to find the specific line of the family containing (1, 2), we substitute 1 for x and 2 for y to determine k.

$$4x-3y=k$$

$$4(1)- 3(2)=k$$

$$-2 = k$$

Thus, the equation for the desired line is $$4x-3y=-2$$

#### Solution

(a) If $$\displaystyle \frac{A}{A'} = \frac{B}{B'} \neq \frac{C}{C'}$$, then the lines are parallel.

Above properties provide us a way to write the equation of a line parallel or perpendicular to a given line that contains a given point not on the line. For example, suppose that we want the equation of the line pependicular to $$3x+4y=6$$ that contains the point (1, 2). The form $$4x-3y=k$$, where k is a constant, represents a family of lines perpendicular to $$3x+4y=6$$ because we have satisfied the condition $$AA' = - BB'$$. Therefore, to find the specific line of the family containing (1, 2), we substitute 1 for x and 2 for y to determine k.

$$4x-3y=k$$

$$4(1)- 3(2)=k$$

$$-2 = k$$

Thus, the equation for the desired line is $$4x-3y=-2$$