CAT Practice: Coordinate Geometry - Free MCQ Test with solutions for Quant

From curve C₁ we can see that x ≥ 0 and y can be either positive or negative. Thus either intersection happens in the first quadrant or fourth quadrant.
x^1/2 = |y| or x = y² ...(I)) from C₁
Putting this in the C₂ equation
y²(y) = -4y² + y
On rearranging we get y³ + 4y² - y  = 0
y(y² + 4y - 1) = 0
Discriminate of (y² + 4y - 1) is positive so it must have 2 distinct roots ≠ 0
y(y² + 4y - 1) = 0 has 3 distinct roots. 
For each value of y there will be unique x such that x^1/2 = |Y|.  

The coordinates of the rectangle are (3,2),(9,2),(3,6) and (9,6).
The line dividing it in two congruent parts should pass through the central point of the rectangle (3 + 9)/2 = 6 and (2 + 6)/2 = 4 i.e (6,4)
The slope of the line passing through (3,0) and (6,4) = 4/3 = a/b
a + b = 4 + 3 = 7

By definition the lines max, (x, y) = 1 means x = 1 and y £ 1 or y = 1 and x £ 1

Required area is shaded area as shown in the figure

= Area of square - Area of semicircle = 1 - (π/4) sq. units

Co-ordinates of D is (12,17)

Let OA = a and OB = b

OAB is a right angled triangle, with AB as hypotenuse

Since AB is a side of the sqaure ABCD, remaining lengths will be as shown in the diagram

Therefore, D’s co-ordinates will be of the form (a, a + b) = (12,17)

a = 12 and b = 5; hence C is (17, 5)

In a parallelogram, two diagonals of parallelogram bisects each other, which concludes that mid-point of both diagonals are the same.

x = -4 and y = 5

The answer is option D.

This is the equation of a circle with radius 4 units and centered at (-2, 3)
From a point L we drop two tangents on the circle such that the angle between the tangents is 60^o.

∠LPO = 90^∘ 
∠PLO=30^∘
 
Therefore the locus of the point L, is a circle centered at (-2, 3) and has a radius of (4 + x = 8) units.
The equation of this locus is thus, (x + 2)2 +(y − 3)2 = 82
 When x = 6, we have, (8)² +(y − 3)² = 8²   , that is y = 3
The circle , (x + 2)² +(y − 3)² = 8², touches the line x = 6 at (6, 3).

So, every point on the line y = 2x − 7 where x ≤ 5 satisfies the given condition.
We are to find the area enclosed by the y-axis, x = 5 and the lines of |x−y|−|x−5|=2 .
Because the area we are interested is bounded by x = 0 (y-axis) and x = 5, 0 ≤ x ≤ 5.
So, we’ll only be concerned about Case III and Case IV.
A rough sketch of the bounded region looks like…

The line y = 2x – 7 touches x = 0 and x = 5 at (0, -7) and (5, 3) respectively.

Given,
The three vertices are: (1,2), (7,2) and (1,10)
The three side lengths are 6, 8 and 10 respectively.
These are the sides of a right angled triangle.
So, Area = 1/2 × 6 × 8 = 24 sq. units
Inradius = area / semi-perimeter = 24 / 12 = 2 units

Given (P(-3,-2)), (Q(1,-5)), and (R(9,1)).
For parallelogram (PQRS), S = P + R - Q = (-3,-2) + (9,1) - (1,-5) = (5,4)
Diagonal SQ passes through S(5,4) and Q(1,-5).