## Question

### Solution

Correct option is

(3, 1)

Old New Pt.

x = X + h = 1 + 2 = 3

y = Y + k = 3 – 2 = 1

(x, y) = (3, 1)

#### SIMILAR QUESTIONS

Q1

A(p, 0), B(4, 0), C(5, 6) and D(1, 4) are the vertices of a quadrilateral ABCD If is obtuse, the maximum integral value of p is

Q2

The value of p for which is obtuse and is a right angle is

Q3

A(p, 0), B(4, 0), C(5, 6) and D(1, 4) are the vertices of a quadrilateral ABCD

If two sides of the quadrilateral are equal; area of the quadrilateral is

Q4

(0, 0), A(1, 1), B(0, 3) are the vertices of a triangle OAB.divides OB in the ratio 1 : 2, Q is the mid-point of AP, R divides AB in the ratio 2 : 1

1:- If Q5

(0, 0), A(1, 1), B(0, 3) are the vertices of a triangle OAB.divides OB in the ratio 1 : 2, Q is the mid-point of AP, R divides AB in the ratio 2 : 1

area of ΔPQR : area of ΔOAB is

Q6

(0, 0), A(1, 1), B(0, 3) are the vertices of a triangle OAB.divides OB in the ratio 1 : 2, Q is the mid-point of AP, R divides AB in the ratio 2 : 1

If S is the mid-point of PR, then QS is equal to

Q7

a and b real numbers between 0 and 1 A(a, 1), B(1, b) andC(0, 0) are the vertices of triangle

1:- If the triangle ABC is equilateral, its area is equal to

Q8

If the triangle ABC in  isosceles with AC = BC and 5(AB)2 = 2(AC)2 then

Q9

The origin is shifted to(1, –2)then what are the coordinates be shifted if the point (3, –5) in the new position?

Q10

Determiner as to what point the axes of the coordinates be shifted so as to remove the first degree terms from the equation

(x, y) = 2x2 + 3y2 – 12x + 12+ 24 = 0