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Chapter 2: Problem 28

Find a number \(t\) such that the point \(\left(t, \frac{t}{2}\right)\) is on the line containing the points (2,-4) and (-3,-11) .

Short Answer

Expert verified
The value of \(t\) such that the point \(\left(t, \frac{t}{2}\right)\) is on the line containing the points (2,-4) and (-3,-11) is \(\boxed{\frac{68}{9}}\).

Step by step solution

01

Find the equation of the line containing the points (2,-4) and (-3,-11)

Given two points, we can find the equation of the line using the slope-point form, which is given by: \[y - y_1 = m(x - x_1)\] where \(m\) is the slope of the line, and \((x_1, y_1)\) is one of the given points. To find the slope, we can use the formula: \[m = \frac{y_2 - y_1}{x_2 - x_1}\] Take the points (2,-4) and (-3,-11) as \((x_1, y_1)\) and \((x_2, y_2)\), respectively. Next, find the slope: \[m = \frac{-11 - (-4)}{-3 - 2} = \frac{-7}{-5} = \frac{7}{5}\] Now, using the slope-point form, plug in the slope and one of the points (let's use the point (2,-4)) to find the equation of the line: \[y - (-4) = \frac{7}{5}(x - 2)\]
02

Substitute the given point and solve for t

We need to find the value of \(t\) such that the point \(\left(t, \frac{t}{2}\right)\) lies on the line found in Step 1. Substitute the point into the equation: \[\frac{t}{2} + 4 = \frac{7}{5}(t - 2)\] Now, solve for \(t\): \[\frac{t}{2} + 4 = \frac{7t}{5} - \frac{14}{5}\] First, multiply both sides by 10 to eliminate the fractions: \[5t + 40 = 14t - 28\] Next, move all the t terms to one side and constant terms to the other side: \[14t - 5t = 40 + 28\] \[9t = 68\] Now, divide by 9: \[t = \frac{68}{9}\] So, the value of \(t\) such that the point \(\left(t, \frac{t}{2}\right)\) is on the line containing the points (2,-4) and (-3,-11) is \(\boxed{\frac{68}{9}}\).

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Most popular questions from this chapter

Suppose \(p\) is a polynomial and \(t\) is a number. Explain why there is a polynomial \(G\) such that $$ \frac{p(x)-p(t)}{x-t}=G(x) $$ for every number \(x \neq t\).

Suppose \(M\) and \(N\) are odd integers. Explain why $$ x^{2}+M x+N $$ has no integer zeros.

Find a polynomial \(p\) of degree 3 such that \(-1,2,\) and 3 are zeros of \(p\) and \(p(0)=1\).

$$ \text { Suppose } p(x)=2 x^{6}+3 x^{5}+5 $$ (a) Show that if \(\frac{M}{N}\) is a zero of \(p\), then $$ 2 M^{6}+3 M^{5} N+5 N^{6}=0 $$ (b) Show that if \(M\) and \(N\) are integers with no common factors and \(\frac{M}{N}\) is a zero of \(p\), then \(5 / M\) and \(2 / N\) are integers. (c) Show that the only possible rational zeros of \(p\) $$ \text { are }-5,-1,-\frac{1}{2}, \text { and }-\frac{5}{2} \text { . } $$ (d) Show that no rational number is a zero of \(p\).

Write the indicated expression as a ratio of polynomials, assuming that $$ r(x)=\frac{3 x+4}{x^{2}+1}, \quad s(x)=\frac{x^{2}+2}{2 x-1}, \quad t(x)=\frac{5}{4 x^{3}+3} $$. $$ (s-t)(x) $$
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