Question

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

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

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: $$y = -4 - \frac{5(\frac{3(x-1)}{2})}{2}$$

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 \(-\frac{15}{4}\).

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.

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.