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xz_007 [3.2K]
3 years ago
14

Write equations for the horizontal and vertical lines passing through the point (8,5)

Mathematics
1 answer:
almond37 [142]3 years ago
7 0

Answers:

Horizontal Line:  y = 5

Vertical Line:  x = 8

=================================================

Explanation:

All horizontal lines are of the form y = k, for some constant k. We want the horizontal line to pass through (8,5), meaning every point on this horizontal line must have y coordinate 5. Therefore, y = 5 is the equation of the horizontal line. Two such points on this line are (1, 5) and (8, 5). All that matters is the y coordinate is 5. The x coordinate can be anything you want. The slope of any horizontal line is 0.

Flipping things around, all vertical lines will have the x coordinate of each point be the same value. Draw a vertical line through (8,5) and note how each point has x coordinate of 8. Two such points are (8,1) and (8,5). Therefore, the equation of the vertical line is x = 8. The y coordinate can be any value you want. The slope of any vertical line is undefined. Unlike the horizontal line, we cannot write this equation in slope intercept form (namely because the slope isn't defined).

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3 years ago
A piece of wire 19 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral tria
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Step-by-step explanation:

$$Let s be the length of the wire used for the square. \\Let $t$ be the length of the wire used for the triangle. \\Let $A_{S}$ be the area of the square. \\Let ${A}_{T}}$ be the area of the triangle. \\One side of the square is $\frac{s}{4}$ \\Therefore,we know that,$$A_{S}=\left(\frac{s}{4}\right)^{2}=\frac{s^{2}}{16}$$

$$The formula for the area of an equilateral triangle is, $A=\frac{\sqrt{3}}{4} a^{2}$ where $a$ is the length of one side,And one side of our triangle is $\frac{t}{3}$So,We know that,$$A_{T}=\frac{\sqrt{3}}{4}\left(\frac{t}{3}\right)^{2}$$We have to find the value of "s" such that,$\mathrm{s}+\mathrm{t}=19$ hence, $\mathrm{t}=19-\mathrm{s}$And$$A_{S}+A_{T}=A_{S+T}$$

$$Therefore,$$\begin{aligned}&A_{T}=\frac{\sqrt{3}}{4}\left(\frac{(19-s)}{3}\right)^{2}=\frac{\sqrt{3}(19-s)^{2}}{36} \\&A_{T+S}=\frac{s^{2}}{16}+\frac{\sqrt{3}(19-s)^{2}}{36}\end{aligned}

$$Differentiating the above equation with respect to s we get,$$A^{\prime}{ }_{T+S}=\frac{s}{8}-\frac{\sqrt{3}(19-s)}{18}$$Now we solve $A_{S+T}^{\prime}=0$$$\begin{aligned}&\Rightarrow \frac{s}{8}-\frac{\sqrt{3}(19-s)}{18}=0 \\&\Rightarrow \frac{s}{8}=\frac{\sqrt{3}(19-s)}{18}\end{aligned}$$Cross multiply,$$\begin{aligned}&18 s=8 \sqrt{3}(19-s) \\&18 s=152 \sqrt{3}-8 \sqrt{3} s \\&(18+8 \sqrt{3}) s=152 \sqrt{3} \\&s=\frac{152 \sqrt{3}}{(18+8 \sqrt{3})} \approx 8.26\end{aligned}$$

$$The domain of $s$ is $[0,19]$.So the endpoints are 0 and 19$$\begin{aligned}&A_{T+S}(0)=\frac{0^{2}}{16}+\frac{\sqrt{3}(19-0)^{2}}{36} \approx 17.36 \\&A_{T+S}(8.26)=\frac{8.26^{2}}{16}+\frac{\sqrt{3}(19-8.26)^{2}}{36} \approx 9.81 \\&A_{T+S}(19)=\frac{19^{2}}{16}+\frac{\sqrt{3}(19-19)^{2}}{36}=22.56\end{aligned}$$

$$Therefore, for the minimum area, $8.26 \mathrm{~m}$ should be used for the square

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