Test: Normal And General Form Of A Line - Commerce MCQ

# Test: Normal And General Form Of A Line - Commerce MCQ

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## 10 Questions MCQ Test Mathematics (Maths) Class 11 - Test: Normal And General Form Of A Line

Test: Normal And General Form Of A Line for Commerce 2024 is part of Mathematics (Maths) Class 11 preparation. The Test: Normal And General Form Of A Line questions and answers have been prepared according to the Commerce exam syllabus.The Test: Normal And General Form Of A Line MCQs are made for Commerce 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Normal And General Form Of A Line below.
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Test: Normal And General Form Of A Line - Question 1

### General equation of a straight line is:

Detailed Solution for Test: Normal And General Form Of A Line - Question 1

The equation Ax + By + c = 0 is the most general equation for a straight line, and can be used where other forms of equation are not suitable.

Test: Normal And General Form Of A Line - Question 2

### The line through the point (a, b) and parallel to the line Ax + By + C = 0 is:

Detailed Solution for Test: Normal And General Form Of A Line - Question 2

Equation of line passing through point (a,b) and parallel to line Ax + By + C = 0
(y-y1) = -A/B(x-x1)
(y-b) = -A/B(x-a)
A (x – a) + B (y – b) = 0

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Test: Normal And General Form Of A Line - Question 3

### Find the perpendicular distance from the origin of the line x + y – 2 = 0 is:

Detailed Solution for Test: Normal And General Form Of A Line - Question 3

The given point is P(0,0) and the given line is x + y - 2 = 0
Let d be the length of the perpendicular from P(0,0) to the line x + y - 2 = 0
Then,
d = |(1 × 0) + (3 × 0) − 2|/(√12 + 12)
= 2/√2
= (2/√2) * (√2/√2)
= √2

Test: Normal And General Form Of A Line - Question 4

The equation of a line whose perpendicular distance from the origin is 8 units and the angle made by perpendicular with positive x-axis is 60 degree is:

Detailed Solution for Test: Normal And General Form Of A Line - Question 4

If p isthe length of the normal from the origin to a line and ωis the angle made by the normal with the positive direction of thex-axis,then the equation of the line is given by xcosω +ysinω= p.
Here, p = 8 units and ω= 60°
Thus, therequired equation of the given line is
xcos 60° + y sin 60° = 8
x(1/2) + y(√3/2) = 8
x/2 + √3y/2 = 8
x + √3y = 16

Test: Normal And General Form Of A Line - Question 5

Equation of line parallel to the line Ax + By + C = 0 is:

Detailed Solution for Test: Normal And General Form Of A Line - Question 5

Let, ax + by + c = 0 (b ≠ 0) be the equation of the given straight line.
Now, convert the equation ax + by + c = 0 to its slope-intercept form.
ax + by+ c = 0
⇒ by = - ax - c
Dividing both sides by b, [b ≠ 0] we get,    y =  -a/b x - cb, which is the slope-intercept form.
Now comparing the above equation to slope-intercept form (y = mx + b) we get,
The slope of the line ax + by + c = 0 is (- a/b).
Since the required line is parallel to the given line, the slope of the required line is also (- ab).
Let k (an arbitrary constant) be the intercept of the required straight line. Then the equation of the straight line is
y = - a/b x + k
⇒ by = - ax + bk
⇒ ax +  by = -K, Where K = bk = another arbitrary constant.

Test: Normal And General Form Of A Line - Question 6

The equation of the line parallel to the line 2x – 3y = 1 and passing through the middle point of the line segment joining the points (1, 3) and (1, –7), is:

Detailed Solution for Test: Normal And General Form Of A Line - Question 6

The midpoint of the line segment is (1+1/2, 3-7/2)
= (1,-2)
the equation of the line parallel to the line 2x-3y = 1 is of the form 2x-3y = k
since it passes through (1,-2)
2(1) - 3(-2) = k
k = 8
hence the required equation is 2x-3y=8

Test: Normal And General Form Of A Line - Question 7

The equation of a line whose perpendicular distance from the origin is 3 units and the angle made by perpendicular with the positive x-axis is 30° is:

Detailed Solution for Test: Normal And General Form Of A Line - Question 7

xcos 30° + y sin 30° = 3
⇒ √3x + y - 6 = 0

Test: Normal And General Form Of A Line - Question 8

Find the equation of the line parallel to the line 5x – 4y + 3 = 0 and passing through the point (2, 5) is:

Detailed Solution for Test: Normal And General Form Of A Line - Question 8

5x – 4y + 3 = 0
5x – 4y = -3
(5/4)x + (-4/5)y = -3
Line is parallel m = 5/4
Points (2,5)
= 5 = (5/4)(2) + b
5 = 5/2 + b
b = 5/2
y = 5x/4 - 5/2
4y = 5x - 10
5x - 4y + 10 = 0

Test: Normal And General Form Of A Line - Question 9

The value of θ, if the equation x cos θ + y sin θ = π is the normal form of the line  √3x + y +2 =0 is:

Detailed Solution for Test: Normal And General Form Of A Line - Question 9

When we form the equation it would be like this:
√3x + y + 2 = 0
There is another way to write this equation: √3x − y = 2
When we divide both sides by 2 we will get this
√3x/2 − 1/2y = 1
Cosθ = √3/2 and sinθ = 1/2, and p=1
We know that both cosθ and sinθ are negatives, so we will get this:
θ = π + π/6
= 7π/6

Test: Normal And General Form Of A Line - Question 10

The equation of straight line in normal form is

Detailed Solution for Test: Normal And General Form Of A Line - Question 10

x Cos ø + y Sin ø = p.
This is called the normal form of equation of the given line making the angle ø with the positive direction of x-axis and whose perpendicular distance from the origin is p.

## Mathematics (Maths) Class 11

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## Mathematics (Maths) Class 11

75 videos|238 docs|91 tests