Question

find a general form of an equation of the line through the point a that satisfies the given condition. a(-6, 3) (a) parallel to the x - axis (b) perpendicular to the x - axis

Explanation:

Step1: Recall line - equation form

The general form of a line is $Ax + By+C = 0$. For a line parallel or perpendicular to the $x$-axis, we can start from the slope - intercept form $y=mx + b$ (where $m$ is the slope and $b$ is the $y$-intercept) and convert it to general form.

Step2: Find equation parallel to x - axis

A line parallel to the $x$-axis has a slope $m = 0$. The line passes through the point $A(-6,3)$. Using the point - slope form $y - y_1=m(x - x_1)$ (where $(x_1,y_1)=(-6,3)$ and $m = 0$), we have $y-3=0(x + 6)$, which simplifies to $y=3$. In general form, it is $y-3=0$.

Step3: Find equation perpendicular to x - axis

A line perpendicular to the $x$-axis has an undefined slope. The line passes through the point $A(-6,3)$. The equation of the line is $x=-6$. In general form, it is $x + 6=0$.

Answer:

(a) $y - 3=0$
(b) $x + 6=0$