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Chapter 2: Problem 16

Find a number \(t\) such that the point \((-2, t)\) is on the line containing the points (5,-2) and (10,-8) .

Short Answer

Expert verified
The value of \(t\) is \(\frac{32}{5}\).

Step by step solution

01

Find the slope of the line.

Given two points (x1, y1) and (x2, y2) on a line, the slope (m) of the line can be found using the formula: \[m = \frac{y2 - y1}{x2 - x1}\] The given points are (5, -2) and (10, -8). Let (x1, y1) = (5, -2) and (x2, y2) = (10, -8). Plug these values into the formula to find the slope: \[m = \frac{-8 - (-2)}{10 - 5}\]
02

Calculate the slope value.

Now, we will simplify the expression to find the slope (m): \[m = \frac{-8 + 2}{5}\] \[m = \frac{-6}{5}\]
03

Use the point-slope form to find the equation of the line.

The point-slope form of a line with slope m and passing through a point (x1, y1) is given by: \[y - y1 = m(x - x1)\] Now, we will use the slope m = -6/5, and let's use the point (5, -2) as (x1, y1). Plugging these into the formula, we get: \[y - (-2) = \frac{-6}{5}(x - 5)\]
04

Simplify the equation of the line.

Now, we will simplify the equation to get the equation of the line in slope-intercept form (y = mx + b): \[y + 2 = \frac{-6}{5}(x - 5)\] \[y = \frac{-6}{5}x + \frac{30}{5} - 2\] \[y = \frac{-6}{5}x + 4\]
05

Substitute the x-coordinate of the point (-2, t) into the equation.

We are given the point (-2, t) and need to find the value of t. Substitute x = -2 into the equation of the line: \[t = \frac{-6}{5}(-2) + 4\]
06

Solve for t.

Now, we will simplify this expression to find the value of t: \[t = \frac{12}{5} + 4\] \[t = \frac{12}{5} + \frac{20}{5}\] \[t = \frac{32}{5}\] Since \( t = \frac{32}{5} \), the point (-2, 32/5) is on the line containing the points (5, -2) and (10, -8).

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Most popular questions from this chapter

Write the indicated expression as \(a\) polynomial. $$ \frac{q(2+x)-q(2)}{x} $$

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Suppose \(q(x)=2 x^{3}-3 x+1\) (a) Show that the point (2,11) is on the graph of \(q\). (b) Show that the slope of a line containing (2,11) and a point on the graph of \(q\) very close to (2,11) is approximately 21 . [Hint: Use the result of Exercise \(17 .]\)
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