Find the equation of the line through (8,-2) perpendicular...

Find the equation of the line through $(8,-2)$ perpendicular to (a) $y={ }_5^4 x+2 ;$ (b) $x+3 y=6 ;$ (c) $x=7$. (a) Any line perpendicular to the given line will have slope $m$ satisfying $\frac{4}{5} m=-1$; thus, $m=-\frac{5}{4}$. The equation of a line through $(8,-2)$ with slope $-\frac{5}{4}$ is found from the point-slope form to be $y-(-2)=-\frac{5}{4}(x-8)$. Simplifying yields $5 x+4 y=32$. (b) First, determine the slope of the given line. Isolating the variable $y$, the equation is seen to be equivalent to $y=-\frac{1}{3} x+2$; hence the slope is $-\frac{1}{3}$. Any line perpendicular to the given line will have slope $m$ satisfying $-\frac{1}{3} m=-1$; thus, $m=3$. The equation of a line through $(8,-2)$ with slope 3 is found from the point-slope form to be $y-(-2)=3(x-8)$. Simplifying yields $y=3 x-26$. (c) Since the given line is vertical, any line perpendicular to the given line must be horizontal, hence must have an equation of the form $y=k$. In this case, $k=-2$; hence $y=-2$ is the required equation.

Find the equation of the line through $(8,-2)$ perpendicular to (a) $y={ }_5^4 x+2 ;$ (b) $x+3 y=6 ;$ (c) $x=7$.
(a) Any line perpendicular to the given line will have slope $m$ satisfying $\frac{4}{5} m=-1$; thus, $m=-\frac{5}{4}$. The equation of a line through $(8,-2)$ with slope $-\frac{5}{4}$ is found from the point-slope form to be $y-(-2)=-\frac{5}{4}(x-8)$. Simplifying yields $5 x+4 y=32$.
(b) First, determine the slope of the given line. Isolating the variable $y$, the equation is seen to be equivalent to $y=-\frac{1}{3} x+2$; hence the slope is $-\frac{1}{3}$. Any line perpendicular to the given line will have slope $m$ satisfying $-\frac{1}{3} m=-1$; thus, $m=3$. The equation of a line through $(8,-2)$ with slope 3 is found from the point-slope form to be $y-(-2)=3(x-8)$. Simplifying yields $y=3 x-26$.
(c) Since the given line is vertical, any line perpendicular to the given line must be horizontal, hence must have an equation of the form $y=k$. In this case, $k=-2$; hence $y=-2$ is the required equation.

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Key Concepts

Slope

The slope is a measure of the steepness or incline of a line and is calculated as the ratio of the vertical change (rise) to the horizontal change (run) between any two points on the line. It is fundamental in determining the direction and angle of a line in the coordinate plane.

Perpendicular Lines

Two lines are perpendicular if the product of their slopes is -1. This means the slope of one line is the negative reciprocal of the slope of the other. This concept is key when finding the equation of a line that must meet another line at a right angle.

Point-Slope Form

The point-slope form of a line, expressed as y - y? = m(x - x?), uses a known point on the line and the slope to define the line’s equation. It is a versatile tool for writing the equation of any line when the slope and a single point are given.

Vertical and Horizontal Lines

Vertical and horizontal lines are special cases in coordinate geometry. Vertical lines have an undefined slope and are represented by an equation of the form x = constant, while horizontal lines have a zero slope and are represented by an equation of the form y = constant. Recognizing these cases is important when determining relationships such as perpendicularity between lines.

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