Problem 81 - Calculus Early Transcendentals

Question

Consider the following pairs of lines. Determine whether the lines are parallel or intersecting. If the lines intersect, then determine the point of intersection. a. \(x=1+s, y=2 s\) and \(x=1+2 t, y=3 t\) b. \(x=2+5 s, y=1+s\) and \(x=4+10 t, y=3+2 t\) c. \(x=1+3 s, y=4+2 s\) and \(x=4-3 t, y=6+4 t\)

Step-by-Step Solution

Answer

Based on the given information, for each pair of lines: a. The lines are not parallel and intersect at point (5, 6). b. The lines are parallel and do not intersect. c. The lines are not parallel and intersect at point (4, 6).

Step 1: Check if the direction vectors are parallel

For lines \(x=1+s, y=2 s\) and \(x=1+2t, y=3t\), the direction vectors are \((1, 2)\) and \((2, 3)\). The lines are parallel if the direction vectors are proportional. Since \(\frac{2}{1} \neq \frac{3}{2}\), the direction vectors are not parallel, and thus, the lines are not parallel.

Step 2: Solve for the intersection point

Since the lines are not parallel, they must intersect. We can find the intersection point by setting \(x\) and \(y\) equal in both equations: \((1+s) = (1+2t)\) and \((2s) = (3t)\) Solve the first equation for \(s\): \(s=2t-1\). Substitute this expression into the second equation: \(2(2t-1)=3t \implies t=2\). Now, substitute \(t=2\) back into either line's equation to find the intersection point: \(x=1+2(2)=5\), \(y=3(2)=6\). Thus, the point of intersection is \((5, 6)\). #b. Pair (b) #

Step 1: Check if the direction vectors are parallel

For lines \(x=2+5s, y=1+s\) and \(x=4+10t, y=3+2t\), the direction vectors are \((5, 1)\) and \((10, 2)\). The lines are parallel if the direction vectors are proportional. Since \(\frac{10}{5} = \frac{2}{1}\), the direction vectors are parallel, and thus, the lines are parallel.

Step 2: No intersection point

As the lines are parallel, there is no intersection point. #c. Pair (c) #

Step 1: Check if the direction vectors are parallel

For lines \(x=1+3s, y=4+2s\) and \(x=4-3t, y=6+4t\), the direction vectors are \((3, 2)\) and \((-3, 4)\). The lines are parallel if the direction vectors are proportional. Since \(\frac{-3}{3} \neq \frac{4}{2}\), the direction vectors are not parallel, and thus, the lines are not parallel.

Step 2: Solve for the intersection point

Since the lines are not parallel, they must intersect. We can find the intersection point by setting \(x\) and \(y\) equal in both equations: \((1+3s) = (4-3t)\) and \((4+2s) = (6+4t)\) Solve the first equation for \(s\): \(s = -t+1\). Substitute this expression into the second equation: \(4+2(-t+1)=6+4t \implies t=1\). Now, substitute \(t=1\) back into either line's equation to find the intersection point: \(x=1+3(1)=4\), \(y=4+2(1)=6\). Thus, the point of intersection is \((4, 6)\).