The area of the quadrilateral formed by the lines | x | + | y | = 1 is

**A. ** 2

**B. ** 8

**C. ** 4

**D. ** None of these

**Answer : ****Option A**

**Explaination / Solution: **

Equations of the lines are

x0, y0, then x+y=1

x0, y0, then x+y=-1

x0, y0, then x-y = 2

x0 , y0, then x-y=-2

Clearly these lines form a square, whose coordinates are (1,1),(1,-1),(-1,-1),(-1,1)

Hence its area is 4 x [1/2]1 x 1= 2 sq units

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Slope of a line is not defined if the line is

**A. ** parallel to the line x – y

**B. ** parallel to X axis

**C. ** parallel to Y axis

**D. ** parallel to the line x+y

**Answer : ****Option C**

**Explaination / Solution: **

Vertical lines hve undefined slopes. Hence a line which is parallel to Y-axis has undefined slopes.

Vertical lines hve undefined slopes. Hence a line which is parallel to Y-axis has undefined slopes.

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The distance between the parallel lines 3x + 4y + 13 = 0 and 3x + 4y – 13 = 0 is

**A. ** 26/5

**B. ** 26/3

**C. ** 26/4

**D. ** 26

**Answer : ****Option A**

**Explaination / Solution: **

Distance between parallel lines is given by

Now substituting the values we get,

=

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The line passing through ( 1, 1 ) and parallel to the line 2x – 3y + 5 = 0 is

**A. ** 3x – 2y = 1

**B. ** 2x + 3y = 5

**C. ** 3x + 2y = 5

**D. ** 2x – 3y + 1= 0

**Answer : ****Option D**

**Explaination / Solution: **

The required line is parallel to the given line, hence it has same slope

Therefore the equation of the required line is 2x-3y+k=0

Since it passes through (1,1)

2(1) - 3(1) +k = 0

Therefore k=1

Hence the equation of the required line is 2x - 3y + 1=0

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The perpendicular distance of the origin from the line 3x +4y + 1 = 0 is

**A. ** none of these.

**B. ** 1/5

**C. ** 1/2

**D. ** 1

**Answer : ****Option B**

**Explaination / Solution: **

The perpendicular distance from the origin to the line is given by

For the given line c = 1, A = 3 and B = 4 and since it passes through the origin,

Sustituting the values we get,

= 1/5

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The lines y = mx , y + 2x = 0 , y = 2x + λ and y = - mx + λ form a rhombus if m =

**A. ** None of these

**B. ** -2

**C. ** -1

**D. ** 1

**Answer : ****Option B**

**Explaination / Solution: **

Rhombus is a parallelogram in which the opposite sides are equaland parallel.

Therefore the lines y = mx and y = -2x are parallel, similarly y = 2x+ and y = -mx + are parallel.

If two lines are equal, then their slopes are qual

This implies m = -2

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The line x + y – 6 = 0 is the right bisector of the segment [PQ]. If P is the point ( 4, 3 ) , then the point Q is

**A. ** ( 3, 4 )

**B. ** ( 3, 2 )

**C. ** ( 3 , 3 )

**D. ** ( 4 , 4 )

**Answer : ****Option B**

**Explaination / Solution: **

Equation of the line which is perpendicular to the given line x + y = 6 is x - y + k = 0

Since it passes through the point (4,3)

4 - 3 + k = 0

Therefore k = -1

Hence the equation of the line segment PQ is x - y - 1 = 0

On solving these two lines we get x = 7/2 and y = 5/2

This point of intersection is the midpoint of the line segement PQ

That is = 7/2 Hence x = 3

Similarly = 5/2. Hence y = 2

Hence the coordiates of the point Q is (3,2)

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If the points representing the complex numbers - 4 +3i , 2 – 3i and 0 + pi are collinear , then the value of p is

**A. ** -1

**B. ** 2

**C. ** 1

**D. ** none of these.

**Answer : ****Option A**

**Explaination / Solution: **

Let us take the coordinates as (-4,3), (2,-3) and (0,p).

If the points are collienear the |x1(y2 - y3) + x2(y3 - y1) + x3(y1-y2)| = 0

Now substituting the values |-4(-3 - p) +2(p -3) +0(3+3)|= 0

12 +4p +2p -6 +0 = 0

6p +6 = 0

6p = -6

Therefore p = -1

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The locus of the inequation xy ≥ 0 is

**A. ** a straight line

**B. ** none of these

**C. ** the set of all points either in the 1st quadrant or in the 3rd quadrant including the points on coordinate axis

**D. ** a pair of straight lines

**Answer : ****Option C**

**Explaination / Solution: **

It is the set of all points either in the 1st quadrant or in the 3rd quadrant including the points on coordiate axis. This is because the inequality indicates that the points belong either to 1st or 3rd quadrant.

It is the set of all points either in the 1st quadrant or in the 3rd quadrant including the points on coordiate axis. This is because the inequality indicates that the points belong either to 1st or 3rd quadrant.

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The lines 8x + 4y = 1, 8x + 4y = 5, 4x + 8y = 3, 4x + 8y = 7 form a

**A. ** Rectangle

**B. ** Square

**C. ** None of these

**D. ** rhombus

**Answer : ****Option D**

**Explaination / Solution: **

On solving the equations 8x+4y=1 and 4x+8y=3, we get the point of intersection as (-1/2,5/12)

On solving the equations 8x+4y=5 and 4x+8y=7, we get the point of intersection as (1/4,3/4)

On solving the equations 8x+4y=1 and 4x+8y=7, weget the point of intersection as (-5/12,13/12)

On solving the equations 8x+4y=5and 4x+8y=3, we the point of intersection as (7/12,1/12)

Let the points A(1/12,5/12), B(7/12,1/12) C(1/4,3/4) and D(-5/12,13/12) be the vertices of the quadrilaeral

Since the slopes of the opposite sides are equal the quadrilateral is a parallelogram

Slope of the diagonal AC is = 1

Slope of the diagonal BD is = -1

Since the product of the slopes is -1, the diagonals are perpendicular to each other

Hence the parallelogram is a rhombus

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