Test Level 2: Coordinate Geometry - 2 - CAT MCQ

Let the points be A(0, 0), B(4, 3), C(3, 5) and  D(-1, 2) 
Let A and B be the easy points then,

The sides are equal. So, the quadrilateral is at least a parallelogram.

and (AC)² ≠ (AB)² + (BC)²
Therefore, the quadrilateral is not rectangle.
So, the given quadrilateral is a parallelogram.

Since, a, b, c are in A.P.
So, 2b = a + c.
Now, only (1, -1) satisfies the equation.

A point will be outside the given circle if its distance from the centre, i.e. origin O, is more than 10.
Now, the distance of point (-7, -8) from O is , which is more than 10.
So, point (-7, -8) lies outside the given circle.
Hence, option (1) is correct.

Slope of the line, m = tan 150^o = tan (90^o + 60^o) = -cot 60^o = -1/√3
y = mx + c 

Line contains the point

The centroid of ΔPQR has coordinates (1/2, 1/3).
Therefore, 1/2 = (x + 0 + 1/2)/3 and 1/3 = (1 + y + 0)/3
So, x = 1 and y = 0
Angle PQR is equal to angle between line PQ and line QR.
Slope of PQ (m1)= (0 - 1)/(0 - 1) = 1
Slope of QR (m2)= (0 - 0)/((1/2) - 0) = 0
Angle between line PQ and line QR = tan θ = (m1 - m2)/(1 + m1m2) = (1 - 0)/(1 + (0)(1)) = 1
tan θ = 1
θ = 45°

The square of the distance from the origin of (x, y) is x² + y².
This values for the different points given in options are 9, (4 + 9), (1 + 9) and 9, respectively. Among these, 4 + 9 is the greatest.

The foot of perpendicular of a point on the x-axis always has abscissa equal to the perpendicular distance of the point from the y-axis and ordinate 0.
The projection or foot of perpendicular of (x, y) on the x-axis is (x, 0).

a² + b² + c² = ab + bc + ac
(a - b)² + (b - c)² + (c - a)² = 0  ...(1)
Since the sum of squares is zero, each term should be zero. 
⇒ (a - b)² = 0, (b - c)² = 0, (c - a)² = 0   
⇒ a = b = c
Hence, the triangle is equilateral.

Since ∠POR = 90° and ∠QOR = 45°, POQ is also equal to 45°.
So, ∠PQO = 45° and ∠QPO = 90°
Therefore, ΔQPO is an isosceles right-angled triangle.
So, OP = PQ = 4 units and OR = 5 units.
Now, OPQR is a trapezium with PQ and OR as its parallel sides.
So, area = ½(PQ + OR) × OP = ½ × (4 + 5) × 4 = 18 sq. units

Since the coordinates of both the points are positive, the line joining them lies in the first quadrant. The division of the line in the ratio of 1 : 1 means that the mid-point of this line also lies in the first quadrant, with both coordinates positive.

(1 + K)X + (3 – K)Y = 2(1 + 3K)            
Check the options.
Option (2): L.H.S. = (1 + K)(5) + (3 – K)(–1)
= 5 + 5K – 3 + K
= 2 + 6K
= 2(1 + 3K) = R.H.S.

The diagonals of a square bisect each other.
So, the point at which the diagonals meet will be the midpoint of a line joining a pair of opposite vertices.
As can be seen from the figure, (-1, -2) and (2, 1) are opposite vertices.
Their midpoint has coordinates ((-1 + 2)/2, (-2 + 1)/2), or (1/2, -1/2).

x–coordinates of R and T are 1 and c, respectively. So, the length of RT = (c – 1).
Length of the altitude from S = Its distance from the x–axis = Its y–coordinate = b 
So, the area of ΔRST = b(c – 1)/2 

Point C(x₃ ,y₃) is the intersection of 3x + 4y = 48 with y = 0.  
So, the coordinates of C are (16, 0).
Point A(x₁ ,y₁) is the intersection of 3y - 4x = 36 with y = 0. 
So, the coordinates of A are (-9, 0).
Point B(x₂ ,y₂) is the intersection of 3x + 4y = 48 and 3y - 4x = 36.
So, the coordinates of B are (0, 12).

Coordinates of the incentre 


= (1, 5)

The equation of the line passing through (a, 0) and (0, b) is: 

On putting the options, we get to know that option (1) is correct.

Coordinates of centroid (G)


y₃ = 5
∴ (x₃, y₃) = (0, 5)

If a line segment with end points A(x₁, y₁) and B(x₂, y₂) is divided in the ratio p : q by a point P(x, y).

For given condition

As (t, 1) is located inside the given circle, its distance from the centre, i.e. origin, should be less than the radius, i.e. √10 units.
∴ Distance between (t, 1) and origin 

so, 

Equation of the line (x/2) + (√3y/2) = 12√5 can also be written as: 
x cos 60° + y sin 60° = 12√5 
Comparing this with the normal form of equation, x cos θ + y sin θ = p, we get the inclination as 60° and the perpendicular distance from the origin as 12√5 units.