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Chapter 2: Problem 44

Find a number \(t\) such that the line containing the points \((t,-2)\) and (-3,5) is perpendicular to the line that contains the points (4,7) and (1,11) .

Short Answer

Expert verified
The number \(t\) that makes the line containing points \((t,-2)\) and \((-3,5)\) perpendicular to the line containing the points \((4,7)\) and \((1,11)\) is \(t = -\frac{37}{3}\).

Step by step solution

01

Find the slope of the first line

In order to find the slope of the line containing the points (4,7) and (1,11), we can use the formula: \[\text{Slope} = \frac{y2 - y1}{x2 - x1}\] Using the given points, we have: \[\text{Slope} = \frac{11 - 7}{1 - 4} = \frac{4}{-3}\] The slope of the line containing points (4,7) and (1,11) is -4/3.
02

Find the slope of the perpendicular line

Since we are looking for a line perpendicular to the first line, we need to find the negative reciprocal of the slope of the first line. To do this, we flip the fraction and change the sign. In this case: \[Perpendicular \ Slope = -\frac{-3}{4} = \frac{3}{4}\]
03

Set up the equation to find t

Now we have the slope of the line containing points (t,-2) and (-3,5) which is 3/4. To find t, we will use the slope formula: \[\frac{5 - (-2)}{-3 - t} = \frac{3}{4}\] Now we can solve for t.
04

Solve for t

To solve for t, we can cross multiply and simplify the equation: \[\frac{7}{-3 - t} = \frac{3}{4}\] \[4 \times 7 = 3 \times (-3 - t)\] \[28 = -9 - 3t\] \[37 = -3t\] \[t = -\frac{37}{3}\]
05

State the final answer

The number t that makes the line containing points (t,-2) and (-3,5) perpendicular to the line containing the points (4,7) and (1,11) is \(t = -\frac{37}{3}\).

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