The figure above shows a right triangle with vertices at the origin

Question

https://gmatclub.com/forum/download/file.php?id=55877 The figure above shows a right triangle with vertices at the origin, (5,6) and (k,0). What is the value of k? (A) 19/3 (B) 58/5 (C) 26/3 (D) 61/5 (E) 37/3 2020-07-09 в 18.48.41.png

firas92 · Accepted Answer

The line joining the points (0,0) and (5,6) has slope  \frac{6-0}{5-0}=\frac{6}{5}

So the line joining the points (k,0) and (5,6) must have slope  \frac{-5}{6}  since it is perpendicular to the first line

\frac{6-0}{5-k} = \frac{-5}{6} 
 36=5k-25 
 k=\frac{61}{5}

Answer is  (D)

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yashikaaggarwal · Answer

Distance between coordinate (5,6) and origin = 
D = √25+36 = √61

Distance between coordinate (5,6) and (K,0) =
D2 = √(5-k)^2+6^2

Distance between coordinate (K,0) and origin =
D3 = √k^2

Using Pythagoras Theorem
K^2 = (5-k)^2+6^2 + 61
K^2 = 25+K^2-10k+36+61
10k = 122
K = 122/10 => 61/5

Answer is D

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ArunSharma12 · Answer

length of the arm with points (0,0) and (5,6):  \sqrt{25 + 36} = \sqrt{61} 
length of the arm with points (k,0) and (5,6):  \sqrt{(k-5)^2 + 36}

area of the triangle =  \frac{\sqrt{61}*\sqrt{(k-5)^2 + 36}}{2} ; as it is a right angled triangle
area of the triangle =  \frac{k*6}{2} ; k is base and 6 is height of the triangle

\sqrt{61}*\sqrt{(k-5)^2 + 36} = k*6

61*((k-5)^2 + 36) = k^2*36

61*(k^2 + 25 - 10k + 36) = k^2*36

25k^2 - 10*61k + 61*61 = 0

k = \frac{61*10 ±\sqrt{(61*10)^2 - 100*61^2}}{50}

k = \frac{61*10 }{50} = \frac{61}{5}

QuantMadeEasy · Answer

Step 1: Understanding the question 
We know, slope of two perpendicular lines are negative reciprocal of each other.

Step 2: Calculation 
Applying formula of slope
Slope of line m1 = (6-0)/(5-0) = 6/5
Slope of line m2 = (0-6)/(k-5) = -6/(k-5)

m1 = -1/m2
6/5 = (k-5)/6
36 = 5k - 25
5k = 61
k = 61/5

(D) is correct

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The figure above shows a right triangle with vertices at the origin

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