Endpoint 2020

Geometry Level 3

The line segment A B AB is 2525 2525 long and the line segment D C DC is perpendicular to A B AB . If the coordinates of the endpoint B B are ( x B , y B ) (x_B, y_B) , find ( x B y B ) (x_B - y_B) .


The answer is -1.

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5 solutions

Chew-Seong Cheong
Dec 26, 2020

Let O O be the origin ( 0 , 0 ) (0,0) ; and A E AE and B E BE be parallel to the x x - and y y -axis respectively. Then we note that A B E \triangle ABE and C D O \triangle CDO are similar and C D O \triangle CDO is a 3 3 - 4 4 - 5 5 right triangle. Since A B = 2525 AB=2525 , x B = A E = 4 5 × 2525 = 2020 x_B = AE = \dfrac 45 \times 2525 = 2020 and y B = B E + 506 = 3 5 × 2525 + 506 = 2021 ( Happy New Year ) y_B = BE + 506 = \dfrac 35 \times 2525 + 506 = 2021 \red {(\text{Happy New Year})} and x B y B = 1 x_B - y_B = \boxed {-1} .

@Ossama Ismail , please use the standard lowercase x x and y y and not the uppercase for axis and coordinates. Put line segments A B AB and D C DC and number 2525 2525 in LaTex.

Happy New Year \Large \red{\text{Happy New Year}}

Chew-Seong Cheong - 5 months, 2 weeks ago

Wishing you a Happy New Year with the hope that you will have many blessings in the year to come. Thanks for your brilliant solutions. I am always enjoying your solutions to my problems.

Ossama Ismail - 5 months, 2 weeks ago

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You are welcome.

Chew-Seong Cheong - 5 months, 2 weeks ago
Tom Engelsman
Dec 25, 2020

The line C D CD can be modeled as y = 4 3 x + 2000 y = -\frac{4}{3}x + 2000 and A B AB as y = 3 4 x + 506. y = \frac{3}{4}x + 506. If A B = 2525 |AB| = 2525 , then we can compute x B x_{B} via the Distance Formula:

( x B 0 ) 2 + ( y B 506 ) 2 = ( x B 0 ) 2 + ( 3 4 x B + 506 506 ) 2 = 2525 ; \sqrt{(x_{B}-0)^2 + (y_{B}-506)^2} = \sqrt{(x_{B} - 0)^2 + (\frac{3}{4}x_{B} + 506 - 506)^2} = 2525;

or x B 2 + 9 16 x B 2 = 2525 ; \sqrt{x^{2}_{B} + \frac{9}{16}x^{2}_{B}} = 2525;

or 25 16 x B 2 = 2525 ; \sqrt{\frac{25}{16}x^{2}_{B}} = 2525;

or 5 4 x B = 2525 ; \frac{5}{4}x_{B} = 2525;

or x B = 2020 y B = 3 4 ( 2020 ) + 506 = 2021. x_{B} = 2020 \Rightarrow y_{B} = \frac{3}{4}(2020) + 506 = 2021.

Hence, x B y B = 2020 2021 = 1 . x_{B} - y_{B} = 2020 - 2021 = \boxed{-1}.

We can just solve it using similar triangles see my solution.

Chew-Seong Cheong - 5 months, 2 weeks ago

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Will do, Chew-Seong.....Happy 2021 and to more ingenious Brilliant solutions!!!

tom engelsman - 5 months, 2 weeks ago

A B C D A B C D = 0 ( x 0 y 506 ) ( 0 1500 2000 0 ) = 0 1500 x + 2000 ( y 506 ) = 0 y 506 = 3 x 4 ( 1 ) \begin{aligned} \overrightarrow{AB}\bot \overrightarrow{CD} & \Rightarrow \overrightarrow{AB}\cdot \overrightarrow{CD}=0 \\ & \Rightarrow \left( \begin{matrix} x-0 \\ y-506 \\ \end{matrix} \right)\cdot \left( \begin{matrix} 0-1500 \\ 2000-0 \\ \end{matrix} \right)=0 \\ & \Rightarrow -1500x+2000\left( y-506 \right)=0 \\ & \Rightarrow y-506=\frac{3x}{4} \ \ \ \ \ (1)\\ \end{aligned}

A B = 2525 x 2 + ( y 506 ) 2 = 2525 ( 1 ) x 2 + ( 3 x 4 ) 2 = 2525 2 x > 0 x = 2020 ( 2 ) \begin{aligned} \left| \overrightarrow{AB} \right|=2525 & \Rightarrow \sqrt{{{x}^{2}}+{{\left( y-506 \right)}^{2}}}=2525 \\ & \overset{\left( 1 \right)}{\mathop{\Rightarrow }}\,{{x}^{2}}+{{\left( \frac{3x}{4} \right)}^{2}}={{2525}^{2}} \\ & \overset{x>0}{\mathop{\Rightarrow }}\,x=2020 \ \ \ \ \ (2)\\ \end{aligned}

( 1 ) , ( 2 ) x y = x 3 x 4 506 = x 4 506 = 2020 4 506 = 1 \left( 1 \right),\left( 2 \right)\Rightarrow x-y=x-\frac{3x}{4}-506=\frac{x}{4}-506=\frac{2020}{4}-506=\boxed{-1}

Pop Wong
Jan 1, 2021

Let the orange line be y = 3 4 x + 506 y=\cfrac{3}{4}x+506 , the slope = 3 4 = \cfrac{3}{4} as it is perpendicular to blue line with slope = 4 3 = \cfrac{-4}{3}

x B 2 + ( y B 506 ) 2 = 252 5 2 x B 2 + ( 3 4 x ) 2 = 252 5 2 x B 2 + 9 16 x B 2 = 252 5 2 25 x B 2 = 16 × 252 5 2 5 x B = 4 × 2525 x B = 4 × 505 = 2020 x B y B = 1 4 x B 506 = 505 506 = 1 x_B^2 + (y_B-506)^2 = 2525^2 \\ \implies x_B^2 + (\cfrac{3}{4}x)^2 = 2525^2 \\ \implies x_B^2 + \cfrac{9}{16} x_B^2 = 2525^2 \\ \implies 25 x_B^2 = 16 \times 2525^2 \\ \implies 5 x_B = 4 \times 2525 \\ \implies x_B = 4 \times 505 = 2020 \\ \implies x_B - y_B = \cfrac{1}{4}x_B-506 = 505-506 = \boxed{-1}

Of course, y B = 2021 y_B = 2021 , Happy New Year!

Saya Suka
Dec 25, 2020

Answer
= (4/5)(2525) - [(3/5)(2525) + 506]
= 2020 - 2021
= -1


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