Test Level 1: Coordinate Geometry CAT MCQs & solutions - Free

P (α, β) lies in the 1st quadrant. Image of point (α, β) in the line x + y = 0 lies in IIIrd quadrant, i.e. (- β, - α)

Point B is the y-intercept of y = 2x - 8, and so has coordinates (0, - 8).
Point A is the x-intercept of y = 2x - 8, so we set y = 0 and obtain 0 = 2x - 8
⇒ 2x = 8
⇒ x = 4
A has coordinates (4, 0).

x-form (x, y) = (x² + x, y² - y); For (-2, 5), coordinates of A are (2 , 20).
y-form (x, y) = (x² - 11, y²); For (-5, 2), co-ordinates of B are (14, 4).
Distance between A and B is


AB = 20 units

The slope of the line through (-2, 3) and (-5, -6) is m

⇒ The slope m1 of the required line 

By point-slope form, Y + 3

⇒ X + 3Y + 11 = 0

b₁y = -a₁x - c₁
b₂y = -a₂x - c₂ 
Slope of line 1 (m₁) =  -a₁/b₁ 
Slope of line 2 (m₂) = -a₂/b₂ 
For lines to be perpendicular, m1m2 = -1
(-a₁/b₁)(-a₂/b₂) = -1
a₁a₂ = -b₁b₂
a₁a₂ + b₁b₂ = 0

Let P(X, Y) be the point which divides P₁(1, -4) and P₂(4, 2) externally in the ratio 1 : 4.  
By section formula, X = {1 + (-1/4)(4)}/(1 - ¼) = 0
Y = {-4 + (-1/4) (2)}/(1 - ¼) = -6

If the vertices of the triangle are at A(8, 0), B(0, 14) and O(0, 0), then [OA] ⊥ [OB]; hence, altitudes through A and B meet at O.
So, the orthocentre of the triangle is at (0, 0).

Length AB = 2a
Now, the vertex C will be such that AC = BC = 2a.
If (x, y) are the coordinates of C, then

Or, a² + x² - 2ax + y² = 9a² + x² - 6ax + y² ⇒ 4ax = 8a²
Or, x = 2a

Or, a² + y² = 4a²
⇒ y² = 3a²

Thus, the coordinates of C are (2a, ±a√3).

Let x-intercept = a and y-intercept = 2a
Using intercept form, we get
Equation of line,
(x/a) + (y/2a) = 1
2x + y = 2a
As it passes through a point (3, 4), therefore,
2(3) + 4 = 2a
a = 5
So, the required equation of line is 2x + y = 10.

Since the circles are concentric, so their centres will be the same.
The centre of circle x² + y² - 3x + ky - 5 = 0 is (3/2, -k/2).
The centre of circle 4x² + 4y² - 12x - y - 9 = 0 is (3/2, 1/8).
This implies,

x – 3y + 1 = 0
⇒ 2x – 6y + 2 = 0 ... (i)
2x + 5y – 9 = 0 ... (ii)
(i) - (ii) gives:
-6y - 5y + 2 + 9 = 0
y = 1
Substitute the value of y in equation (i), we get
x = 2
So, the intersection point of x – 3y + 1 = 0 and 2x + 5y – 9 = 0 is (2, 1) and m = 1/0 (because the slope is infinite).
So, the required line is (y – 1) = (1/0)(x – 2), which gives x - 2 = 0 or x = 2.

On checking the options one by one, we get that only option (4) satisfies the given curve.
Putting x = 9 and y = 4√2, we get

Thus, option 4 is correct.

The perpendicular distance of the vertex (3, -5) from the line is the altitude of the triangle.

Required area of triangle 

Equation of a circle = x² + y² + 2gx + 2fy + c = 0
where r = radius of the circle 
Comparing with the given equation, g = -3, f = -4, c = -24

Area of a circle = πr² = 49π sq. units

Since a₁/a₂ = b₁/b₂ = c₁/c₂ , = μ (say), therefore the straight lines u = 0 and v = 0 are coincident.
Also, u + λv = 0.
⇔ a₁x + b₁y + c₁ + λ(a₂x + b₂y + c₂) = 0
⇔ μ(a₂x + b₂y + c₂) + λ(a₂x + b₂y + c₂) = 0
⇔ (μ + λ)(a₂x + b₂y + c₂) = 0
⇔ (μ + λ) v = 0
So, v = 0
Similarly, u = 0