Difference between revisions of "2023 AMC 8 Problems/Problem 7"

Revision as of 07:15, 26 January 2023

Problem

A rectangle, with sides parallel to the $x$ -axis and $y$ -axis, has opposite vertices located at $(15, 3)$ and $(16, 5)$ . A line drawn through points $A(0, 0)$ and $B(3, 1)$ . Another line is drawn through points $C(0, 10)$ and $D(2, 9)$ . How many points on the rectangle lie on at least one of the two lines?

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

Solution 1

If we extend the lines, we have

Hence, we see that the answer is $\boxed{\textbf{(B)}\ 1}$

~MrThinker

Solution 2

If the analytic expression for line $AB$ is $y=k_{1}x+b_{1}$ , and the analytic expression for line $CD$ is $y=k_{2}x+b_{2}$ , we have the systems of equations: $\begin{cases} b_{1} = 0 \\ 3k_{1} + b_{1} = 1 \end{cases}$ and $\begin{cases} b_{2} = 10 \\ 2k_{2} + b_{2} = 9 \end{cases}$ Solving the systems, we have: $\begin{cases} k_{1} = \frac{1}{3} \\ b_{1} = 0 \end{cases}$ and $\begin{cases} k_{2} = -\frac{1}{2} \\ b_{2} = 10 \end{cases}$ Therefore, we can determine that the expression for line $AB$ is $y=\frac{1}{3}x$ . and the expression for line $CD$ is $y=-\frac{1}{2}x + 10$ . When $x=15$ , the coordinates that line $AB$ and line $CD$ pass through are $(15, 5)$ and $(15, \frac{5}{2})$ , and $(15, 5)$ lies perfectly on one vertex of the rectangle while the $y$ coordinate of $(15, \frac{5}{2})$ is out of the range $3 \leq y \leq 5$ (lower than the bottom left corner of the rectangle $(15, 3)$ ). Considering that the $x$ value of the line $CD$ will only decrease, and the $x$ value of the line $AB$ will only increase, there will not be another point on the rectangle that lies on either of the two lines. Thus, we can conclude that the answer is $\boxed{\textbf{(B)}\ 1}$

~Bloggish

Video Solution by Magic Square

https://youtu.be/-N46BeEKaCQ?t=5151

@@ Line 5: / Line 5: @@
 <math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4</math>
-==Solution==
+==Solution 1==
 If we extend the lines, we have
@@ Line 13: / Line 13: @@
 ~MrThinker
+==Solution 2==
+If the analytic expression for line <math> AB </math> is <math> y=k_{1}x+b_{1} </math>, and the analytic expression for line <math> CD </math> is <math> y=k_{2}x+b_{2} </math>, we have the systems of equations:
+<cmath>  \begin{cases} b_{1} = 0 \\ 3k_{1} + b_{1} = 1 \end{cases}  </cmath>
+and
+<cmath>  \begin{cases} b_{2} = 10 \\ 2k_{2} + b_{2} = 9 \end{cases}  </cmath>
+Solving the systems, we have:
+<cmath> \begin{cases} k_{1} = \frac{1}{3} \\ b_{1} = 0 \end{cases} </cmath>
+and
+<cmath> \begin{cases} k_{2} = -\frac{1}{2} \\ b_{2} = 10 \end{cases} </cmath>
+Therefore, we can determine that the expression for line <math> AB </math> is <math> y=\frac{1}{3}x </math>. and the expression for line <math> CD </math> is <math> y=-\frac{1}{2}x + 10 </math>. When <math> x=15 </math>, the coordinates that line <math> AB </math> and line <math> CD </math> pass through are <math> (15, 5) </math> and <math> (15, \frac{5}{2}) </math>, and <math> (15, 5) </math> lies perfectly on one vertex of the rectangle while the <math> y </math> coordinate of <math> (15, \frac{5}{2}) </math> is out of the range <math> 3 \leq y \leq 5 </math> (lower than the bottom left corner of the rectangle <math> (15, 3) </math>). Considering that the <math> x </math> value of the line <math> CD </math> will only decrease, and the <math> x </math> value of the line <math> AB </math> will only increase, there will not be another point on the rectangle that lies on either of the two lines. Thus, we can conclude that the answer is <math>\boxed{\textbf{(B)}\ 1}</math>
+~Bloggish
 ==Video Solution by Magic Square==

2023 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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