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Chapter 2: Problem 56

Show that the points \((-84,-14),(21,1),\) and (98,12) lie on a line.

Short Answer

Expert verified
The slopes between the points \((-84,-14)\) and \((21,1)\), and between the points \((21,1)\) and \((98,12)\) are both equal to \(\frac{1}{7}\). Therefore, the three points lie on the same straight line.

Step by step solution

01

Calculate the slope between the first two points.

We use the formula for the slope of a line given two points: \[m = \frac{y_2 - y_1}{x_2 - x_1}\] So, for the points \((-84,-14)\) and \((21,1)\), we have: \[m_1 = \frac{1 - (-14)}{21 - (-84)}\] Now, calculate the slope between the second and third points, \((21,1)\) and \((98,12)\).
02

Calculate the slope between the second and third points.

Using the same formula for the slope of a line, for points \((21,1)\) and \((98,12)\), we have: \[m_2 = \frac{12 - 1}{98 - 21}\] Now compare the slopes of the line segments to determine if they are equal.
03

Compare the slopes.

\[ m_1 = \frac{1 - (-14)}{21 - (-84)} = \frac{15}{105} \] and \[ m_2 = \frac{12 - 1}{98 - 21} = \frac{11}{77} \] Simplify the fractions: \[ m_1 = \frac{15}{105} = \frac{1}{7} \] and \[ m_2 = \frac{11}{77} = \frac{1}{7} \] We can see that the slopes are equal, i.e., \(m_{1} = m_{2}\).
04

Conclusion

Since the slopes of the line segments are equal, we can conclude that the three points \((-84,-14)\), \((21,1)\), and \((98,12)\) lie on the same straight line.

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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 \(p\) and \(q\) are polynomials and the horizonal axis is an asymptote of the graph of \(\frac{p}{q}\). Explain why $$ \operatorname{deg} p<\operatorname{deg} q $$

Suppose \(t\) is a zero of the polynomial \(p\) defined by $$ p(x)=3 x^{5}+7 x^{4}+2 x+6 $$ Show that \(\frac{1}{t}\) is a zero of the polynomial \(q\) defined by $$ q(x)=3+7 x+2 x^{4}+6 x^{5} $$.

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 \(17 .]\)

Explain why the composition of two polynomials is a polynomial.
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