1. 1st quadrant

2. 2nd quadrant

3. 3rd quadrant.

4. 4th quadrant

The point is negative in the x axis and positive for the y axis , thus the point must lie in the 2nd quadrant.

1. (2, 0) and (10, 0).

2. (0,2) and ( 0,10).

3. (2, 10) and (0, 0).

4. None of these

Let the co-ordinates of the point on the x-axis be (x, 0).

As distance =&redic;(x

so 5 = &redic;(x-6)

⇒ 25 = x2 - 12x + 36 + 9.

⇒ x = 2 or x = 10

So the required points are (2, 0) and (10, 0).

1. 1st & 2nd quadrant

2. 2nd & 3rd quadrant

3. 1st & 3rd quadrant

4.2nd & 4th quadrant

The signs are different for 2nd and 4th quadrants.

1. (5/3 , 1/3)

2. (3/8 , 3/11)

3. (8/3 , 11/3)

4. (5/3 , 17/3)

Point of intersection of medians is called as centroid, at which each median is divided in the ratio 2 : 1. Firstly find the midpoint of NO and take it as P

Take the point C, where medians meet and it will divide the median MP in the ratio 2: 1.

1. 2

2. 1

3. 3

4. 4

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1. y =2/3x- 1

2. Y=3x-2

3. Y=2x-3

4. None of these

Given m = 2 and c = - 3. Substituting values in y = mx + c, we get y = 2x – 3.

1. y=2/5x- 1

2. Y=5x-2

3. Y=2x-5

4. None of these

The given line is 2x – y = 4 ⇒ y = 2x – 4 (Converting into the form of y = mx + c)

Its slope = 2. The slope of the parallel line should also be 2.

Hence for the required line

m = 2 and (x1 , y1) = (2, -1).

Equation = (y-y

⇒ (y-y

⇒ y-y

⇒ y = 2x – 5.

1. (5/3 , 1/3)

2. (3/8 , 3/11)

3. (8/3 , 11/3)

4. (11/3 , 17/3)

The internal division will use the formula (mx

y = (my

So, the point becomes (11/3,17/3).

1. (5/3 , 1/3)

2. (3/49 , 1/10)

3. (49/3 , 10)

4.None of these

The external division case will use the formula

x = (mx

y = (my

where m:n is 5:2 in our case.

Putting the values you will get points (49/3, 10).

1. y =5/3x- 2

2. 3Y=2x+5

3. 3Y=5x-2

4. None of these

The given line is 3x + 2y + 4 = 0 or y = (-3x/2) - 2

Any line perpendicular to it will have slope = 2/3

Thus equation of line through (2, 3) and slope 2/3 is

(y – 3) = 2/3 (x – 2)

⇒ 3y – 9 = 2x – 4

⇒ 3y – 2x – 5 = 0.

1. 3

2. 4.5

3. 4

4. 3.5

The area of the triangle having its vertices as P(x1, y1), Q(x2, y2) and R(x3, y3) is given by