Geometry - Coordinate Geometry - Previous Year CAT/MBA Questions

Answer: Option D

Text Explanation :

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The length of AC = (b + 4c) - (b - 2c) = 6c

Altitude from C to AB = a - (-2a) = 3a

∴ Area of ABC = 1/2 × b × h = 1/2 × 3a × 6c = 9ac

Hence, option (d).

Answer: Option E

Text Explanation :

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Given, BC = 5 and BE = 3

In △ABD ⇒ AB² = AD² + BD²    ...(1)
In △AED ⇒ AE² = AD² + DE²    ...(2)

(1) - (2) 
⇒ AB² - AE² = BD² - DE² 
⇒ AB² - AE² = x² - (3 - x)² 
⇒ AB² - AE² = x² - 9 - x² + 6x 
⇒ AB² - AE² = - 9 + 6x 

We need to find AB² - AE² + 6CD
= (-9 + 6x) + 6 × (5 - x)
= -9 + 6x + 30 - 6x
= 21

Hence, option (e).

Answer: Option E

Text Explanation :

P be the point of intersection of the lines 3x + 4y = 2a and 7x + 2y = 2018.

Solving these two equations we get the coordinates of P in terms of a.

∴ P ≡ $\left(\frac{4036-2a}{11},\frac{7a-3027}{11}\right)$

Q the point of intersection of the lines 3x + 4y = 2018 and 5x + 3y = 1

Solving these two equations we get the coordinates of Q.

∴ Q ≡ (-550, 917)

Now slope of line connecting P and Q is 2

∴ $\frac{{\displaystyle \frac{7a-3027}{11}}-917}{{\displaystyle \frac{4036-2a}{11}}+550}$ = $\frac{2}{1}$

⇒ $\frac{7a-13114}{10086-2a}$ = $\frac{2}{1}$

⇒ 11a = 33286

⇒ a = 3026

Hence, option (e).

Answer: Option D

Text Explanation :

(13.5, 16) and (5.5, 7.5) are adjacent points of the parallelogram and (17.5, 23.5) is the point of intersection of two diagonals of the parallelogram.

Property: Diagonals of a parallelogram bisect each other.

Distance between (17.5, 23.5) and (5.5, 7.5):

$\sqrt{{(17.5-5.5)}^2+{(23.5-7.5)}^2}=20$

Length of diagonal that passes through (17.5, 23.5) and (5.5, 7.5) = 20 × 2 = 40 cm
Distance between (17.5, 23.5) and (13.5, 16):

$\sqrt{{(17.5-13.5)}^2+{(23.5-16)}^2}=8.5$

Length of diagonal that passes through (17.5, 23.5) and (13.5, 16) = 8.5 × 2 = 17 cm

Hence, option (d).