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2023 AMC 8 Problems/Problem 7

Revision as of 19:20, 26 January 2023 by Bloggish (talk | contribs) (→‎Solution 2)

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

Screenshot 2023-01-25 8.26.48 AM.png

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

~MrThinker

Solution 2

Note that the $y$-intercept of line $AB$ and line $CD$ are $0$ and $10$. If the analytic expression for line $AB$ is $y=k_{1}x$, and the analytic expression for line $CD$ is $y=k_{2}x+10$, we have equations:$3k_{1} = 1$ and $2k_{2} + 10 = 9$. Solving these equations, we can find out that $k_{1} = \frac{1}{3}$ and $k_{2} = -\frac{1}{2}$. 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 $\left(15, \frac{5}{2}\right)$, and $(15, 5)$ lies perfectly on one vertex of the rectangle while the $y$ coordinate of $\left(15, \frac{5}{2}\right)$ is out of the range $3 \leq y \leq 5$ (lower than the bottom left corner of the rectangle $(15, 3)$). Considering that the $y$ value of the line $CD$ will only decrease, and the $y$ 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

See Also

2023 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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