Test: Basic Concepts Of 2D Geometry - JEE MCQ

# Test: Basic Concepts Of 2D Geometry - JEE MCQ

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## 10 Questions MCQ Test Chapter-wise Tests for JEE Main & Advanced - Test: Basic Concepts Of 2D Geometry

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Test: Basic Concepts Of 2D Geometry - Question 1

### Find the value of x for which the points (1,3) , (-2, 9) and (x, -1) are collinear.

Detailed Solution for Test: Basic Concepts Of 2D Geometry - Question 1

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

Test: Basic Concepts Of 2D Geometry - Question 2

### The ratio in which the point R (1, 2) divides the line segment joining points P (2, 3) and Q (3, 5) is:

Detailed Solution for Test: Basic Concepts Of 2D Geometry - Question 2

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

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Test: Basic Concepts Of 2D Geometry - Question 3

### The distance of (2,3) from x+y = 1 is

Detailed Solution for Test: Basic Concepts Of 2D Geometry - Question 3

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

Test: Basic Concepts Of 2D Geometry - Question 4

The distance between two points A (5, 6) and B (8, 3) is:

Detailed Solution for Test: Basic Concepts Of 2D Geometry - Question 4

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

Test: Basic Concepts Of 2D Geometry - Question 5

The area of the triangle whose vertices are (2, 3), (3, 5) and (7, 5) is:

Detailed Solution for Test: Basic Concepts Of 2D Geometry - Question 5

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

Test: Basic Concepts Of 2D Geometry - Question 6

The slope of line when coordinates of any two point A (x1, y1) and B (x2, y2) on the line are given is:

Detailed Solution for Test: Basic Concepts Of 2D Geometry - Question 6

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

Test: Basic Concepts Of 2D Geometry - Question 7

____ is the slope or gradient of a line, if q is the inclination of a line from X-axis.

Detailed Solution for Test: Basic Concepts Of 2D Geometry - Question 7

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.

Test: Basic Concepts Of 2D Geometry - Question 8

The coordinate _______ is at 5 units distance from the X-axis measured along the negative Y-axis and has zero distance from the y-axis.

Detailed Solution for Test: Basic Concepts Of 2D Geometry - Question 8

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)

Test: Basic Concepts Of 2D Geometry - Question 9

Find the slope of line passing through the point A (5, 4) and B (4, 6).

Detailed Solution for Test: Basic Concepts Of 2D Geometry - Question 9

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

Test: Basic Concepts Of 2D Geometry - Question 10

Two lines 3x+4y = 8 and lx+my = n are perpendicular. Which of the following is true?

Detailed Solution for Test: Basic Concepts Of 2D Geometry - Question 10

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

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