Basic Concepts Of 2D Geometry - Free MCQ Practice Test with solutions,

Let A(1,3), B(-2,9), and C(x,-1)
For to be points collinear,
x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)=0
⇒ 1(9-(-1)) + (-2)(-1-3) + x(3-9)=0
⇒ 1(10)+(-2)(-4)+x(-6)=0
⇒ 10+8-6x=0
⇒ 18 = 6x
⇒ x = 3

P(2,3) Q(3,5) R(1,2)
R is at centre between P and Q, using section formula for internal division
Therefore, (1,2) = ((3λ+2)/(λ+1), (5λ+3)/(λ+1))
1 = (3λ+2)/(λ+1)
(λ+1) = (3λ+2)
λ = -1/2
- sign indicates the external division

Given circle = (2,3)
Given line (x+y-1) = 0
Distance between point to line is:
d = |ax₁ + by₁ + c|/√(a² + b²)
where a = 1, b = 1, c = -1 and x1 = 2, y1 = 3
d = |1(2) + 1(3) - 1|/√(1+1)
d = 4/(√2)
d = 2√2

Distance between two points in Cartesian system can be found using distance formula,
AB = [(5−8)² + (6−3)²]^1/2
AB = [(-3)² + (3)²]^1/2
AB = (18)^1/2
AB = 3(2)^1/2

Area of triangle = 1/2 × base × height
= 1/2×(7−3)×(5−3)
= 1/2*(4×2)
= 4 sq units.

Let the line make an angle P with the positive direction of the x-axis (see Fig.). Draw AL and BM perpendicular to x-axis and AN perpendicular to BM. Clearly, ∠NAB = θ​
tanθ = BN/AN = BN/LM
= BM-MN/OM-OL
= BM-AL/OM-OL  = y₂-y₁/x₂-x₁

We know that if q is the inclination of a straight line, then tanq is called the slope (or gradient) of a line. The slope of a line is usually denoted by m.

The coordinate has 5 units distance from X axis, so Y-coordinate is ±5.
The coordinate has 0 distance from the Y axis, so X-coordinate is 0.
Coordinate of the points: (0, ±5)

Slope of line passing through two points (x₁ , y₁) and (x₂ , y₂) is given by m = y₂−y₁ / x₂−x₁
Since the line passes through (5,4) and (4,6)
We have
Slope(m) = 6−4/4−5
= − 2

In perpendicular, the slope of product = -1
Slope of L1 * slope of L2 = -1
-3/4 * (-l/m) = -1
=> 3l/4m = -1
=> 3l = -4m
= 3l + 4m = 0