Difference between revisions of "2017 AMC 10B Problems/Problem 10"
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==Problem== | ==Problem== | ||
| − | The lines with equations <math>ax-2y=c</math> and <math>2x+by=-c</math> are perpendicular and intersect at <math>(1, 5)</math>. What is <math>c</math>? | + | The lines with equations <math>ax-2y=c</math> and <math>2x+by=-c</math> are perpendicular and intersect at <math>(1, -5)</math>. What is <math>c</math>? |
<math>\textbf{(A)}\ -13\qquad\textbf{(B)}\ -8\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 13</math> | <math>\textbf{(A)}\ -13\qquad\textbf{(B)}\ -8\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 13</math> | ||
==Solution== | ==Solution== | ||
| − | Writing each equation in slope-intercept form, we get <math>y=\frac{a}{2}x-\frac{1}{2}c</math> and <math>y=-\frac{2}{b}x-\frac{c}{b}</math>. We observe the slope of each equation is <math>\frac{a}{2}</math> and <math>-\frac{2}{b}</math>. | + | Writing each equation in slope-intercept form, we get <math>y=\frac{a}{2}x-\frac{1}{2}c</math> and <math>y=-\frac{2}{b}x-\frac{c}{b}</math>. We observe the slope of each equation is <math>\frac{a}{2}</math> and <math>-\frac{2}{b}</math>, respectively. Because the slope of a line perpendicular to a line with slope <math>m</math> is <math>-\frac{1}{m}</math>, we see that <math>\frac{a}{2}=-\frac{1}{-\frac{2}{b}}</math> because it is given that the two lines are perpendicular. This equation simplifies to <math>a=b</math>. |
| − | <math>\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> |
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/V4t05w7-Zd4 | ||
| + | |||
| + | ~savannahsolver | ||
| + | |||
| + | ==Video Solution by TheBeautyofMath== | ||
| + | https://youtu.be/XRfOULUmWbY?t=582 | ||
| + | |||
| + | ~IceMatrix | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2017|ab=B|num-b=9|num-a=11}} | {{AMC10 box|year=2017|ab=B|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 17:50, 17 January 2021
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 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
