Difference between revisions of "2017 AMC 10B Problems/Problem 10"
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Because <math>(1, -5)</math> is a solution of both equations, we deduce <math>a \times 1-2 \times -5=c</math> and <math>2 \times 1+b \times -5=-c</math>. Because we know that <math>a=b</math>, the equations reduce to <math>a+10=c</math> and <math>2-5a=-c</math>. Solving this system of equations, we get <math>c=\boxed{\textbf{(E)}\ 13}</math> | Because <math>(1, -5)</math> is a solution of both equations, we deduce <math>a \times 1-2 \times -5=c</math> and <math>2 \times 1+b \times -5=-c</math>. Because we know that <math>a=b</math>, the equations reduce to <math>a+10=c</math> and <math>2-5a=-c</math>. Solving this system of equations, we get <math>c=\boxed{\textbf{(E)}\ 13}</math> | ||
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| + | == Solution 2 == | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 19:19, 3 July 2025
Contents
Problem
The lines with equations 
and 
are perpendicular and intersect at 
. What is 
?
Solution
Writing each equation in slope-intercept form, we get 
and 
. We observe the slope of each equation is 
and 
, respectively. Because the slope of a line perpendicular to a line with slope 
is 
, we see that 
because it is given that the two lines are perpendicular. This equation simplifies to 
.
Because is a solution of both equations, we deduce 
and 
. Because we know that , the equations reduce to 
and 
. Solving this system of equations, we get 
Solution 2
Video Solution
~savannahsolver
Video Solution by TheBeautyofMath
https://youtu.be/XRfOULUmWbY?t=582
~IceMatrix
See Also
| 2017 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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 AMC 10 Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
