b) Perpendicular to the line $2 x-4 y=-3$ with the same $x$ -intercept as $y=-5 x+20$ .

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Write an equation for a parallel or perpendicular line

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Q. b) Perpendicular to the line

2 x-4 y=-3

with the same

x

-intercept as

y=-5 x+20

Find Slope: First, let's find the slope of the given line $2x-4y=-3$ . $\newline$ We can rearrange the equation into slope-intercept form ( $y=mx+b$ ) to find the slope.
Negative Reciprocal: The equation $2x-4y=-3$ can be rewritten as $4y=2x+3$ by adding $4y$ to both sides and then dividing by $4$ to isolate $y$ . $\newline$ $y = \frac{2}{4}x + \frac{3}{4}$ $\newline$ $y = \frac{1}{2}x + \frac{3}{4}$ $\newline$ The slope $(m)$ of this line is $\frac{1}{2}$ .
X-Intercept: Since we are looking for a line perpendicular to the given line, we need the negative reciprocal of the slope of the given line. $\newline$ The negative reciprocal of $\frac{1}{2}$ is $-2$ .
Point-Slope Form: Now, let's find the $x$ -intercept of the line $y=-5x+20$ . $\newline$ The $x$ -intercept occurs where $y=0$ .
Equation of Perpendicular Line: Setting $y$ to $0$ in the equation $y=-5x+20$ gives us: $\newline$ $0 = -5x + 20$ $\newline$ Adding $5x$ to both sides gives us: $\newline$ $5x = 20$ $\newline$ Dividing both sides by $5$ gives us: $\newline$ $x = 4$ $\newline$ So the x-intercept is at the point $(4,0)$ .
Equation of Perpendicular Line: Setting $y$ to $0$ in the equation $y=-5x+20$ gives us: $\newline$ $0 = -5x + 20$ $\newline$ Adding $5x$ to both sides gives us: $\newline$ $5x = 20$ $\newline$ Dividing both sides by $5$ gives us: $\newline$ $x = 4$ $\newline$ So the x-intercept is at the point $(4,0)$ .We now have the slope of the line we are looking for, which is $-2$ , and a point it passes through, which is $(4,0)$ . $\newline$ We can use the point-slope form of the equation of a line to find the equation of our line. $\newline$ The point-slope form is $0$ $1$ , where $0$ $2$ is the slope and $0$ $3$ is the point the line passes through.
Equation of Perpendicular Line: Setting $y$ to $0$ in the equation $y=-5x+20$ gives us: $\newline$ $0 = -5x + 20$ $\newline$ Adding $5x$ to both sides gives us: $\newline$ $5x = 20$ $\newline$ Dividing both sides by $5$ gives us: $\newline$ $x = 4$ $\newline$ So the x-intercept is at the point $(4,0)$ .We now have the slope of the line we are looking for, which is $-2$ , and a point it passes through, which is $(4,0)$ . $\newline$ We can use the point-slope form of the equation of a line to find the equation of our line. $\newline$ The point-slope form is $0$ $1$ , where $0$ $2$ is the slope and $0$ $3$ is the point the line passes through.Plugging our slope and point into the point-slope form gives us: $\newline$ $0$ $4$ $\newline$ Simplifying this we get: $\newline$ $0$ $5$ $\newline$ This is the equation of the line perpendicular to $0$ $6$ that has the same x-intercept as $y=-5x+20$ .