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Chapter 11: Problem 36

Find the slope of each line and a point on the line. Then graph the line. $$x=1+2 t / 3, y=-4-5 t / 2$$

Short Answer

Expert verified
Answer: The slope of the line is -15/4, a point on the line is (1, -4). To graph the line, plot the point (1, -4) and use the slope -15/4 to find another point on the line, then draw a straight line through these two points.

Step by step solution

01

Eliminate the parameter t

We have the parametric equations: $$x=1+\frac{2t}{3}$$ $$y=-4-\frac{5t}{2}$$ First, we need to solve one of the equations for \(t\) and then substitute this expression into the other equation to eliminate the parameter \(t\). Let's solve the first equation for \(t\): $$t = \frac{3(x-1)}{2}$$ Now substitute this expression for \(t\) into the second equation: $$y = -4 - \frac{5(\frac{3(x-1)}{2})}{2}$$
02

Simplify the equation in slope-intercept form

Now, we'll simplify this equation to find the equation of the line in slope-intercept form (\(y = mx + b\)): $$y = -4 - \frac{15(x-1)}{4}$$ $$y = -4 - \frac{15x}{4} + \frac{15}{4}$$ $$y = -\frac{15x}{4}+\frac{1}{4}$$ So, the slope-intercept form of the line is \(y = -\frac{15}{4}x + \frac{1}{4}\), where the slope (\(m\)) is \(-\frac{15}{4}\).
03

Determine a point on the line

To find a point on the line, substitute any value of \(t\) into the original parametric equations, and we will get the coordinates of the corresponding point. Let's use \(t=0\): $$x(0) = 1 + \frac{2(0)}{3} = 1$$ $$y(0) = -4 - \frac{5(0)}{2} = -4$$ So, the point \((1, -4)\) is on the line.
04

Graph the line

To graph the line, first plot the point \((1, -4)\). Then, use the slope \(-\frac{15}{4}\) to find another point on the line by moving 15 units down and 4 units to the right from the initial point. Finally, draw a straight line through these two points. Now, we have determined the slope of the line \(-\frac{15}{4}\) and found a point \((1, -4)\) on the line. In addition, we have plotted the line using the slope and a point.

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Most popular questions from this chapter

Use a graphing utility to graph the parabolas \(y^{2}=4 p x,\) for \(p=-5,-2,-1,1,2,\) and 5 on the same set of axes. Explain how the shapes of the curves vary as \(p\) changes.

Find an equation of the line tangent to the hyperbola \(x^{2} / a^{2}-y^{2} / b^{2}=1\) at the point \(\left(x_{0}, y_{0}\right)\)

What is the equation of the standard hyperbola with vertices at \((0, \pm a)\) and foci at \((0, \pm c) ?\)

Give the property that defines all parabolas.

Find an equation of the following parabolas, assuming the vertex is at the origin. A parabola that opens to the right with directrix \(x=-4\)
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