Coordinate Geometry(Distance formula and Section formula)

Distance between P(0,y) and Q(x.0) is given by-

• $x^2+y^2$

• $\sqrt{x^2-y^2}$

• $\sqrt{x^2+y^2}$

• $\sqrt{xy}$

Distance of P(2,3) from x-axis is-

• 2

• 3

• 1

• 5

Distance of P(-6,8) from y-axis is-

• 8

• $2\sqrt{7}$

• 10

• 6

Find a point on x-axis that is equidistant from A(-2,9) and B(2,-5)

• (0,7)

• (-7,0)

• (7,0)

• (7,-7)

 $P\left(\frac{a}{3},4\right)\ is\ the\ midpoint\ of\ line\ segment\ joining\ Q\left(-6,5\right)\ and\ R\left(-2,3\right),\ then\ value\ of\ a\ is-$  

• -4

• -12

• 12

• -6

M(-1,1) is the midpoint of line segment joining P(-3,y) and Q(1,y+4), then y is-

• 1

• -1

• 2

• 0

AOBC is a rectangle with three vertices A(0,3), O(0,0) and B(5,0). then the length of diagonal is

• 5

• 3

• $\sqrt{34}$

• 4

 $Distnace\ between\ the\ points\ \left(a\cos\theta+b\sin\theta,0\right)\ and\ \left(0,\ a\sin\theta-b\cos\theta\right)is-$  

•  $a^2+b^2$  

• a+b

•  $a^2-b^2$  

•  $\sqrt{a^2+b^2}$  

P(a, -11), Q(5,b), R(2,15) and S(1,1) are vertices of a parallelogram PQRS taken in order. then value of a and b are-

• a = 4, b = 4

• a = 4, b = 2

• a = 3, b = 4

• a = 4, b = 3

 $Find\ the\ third\ vertex\ of\ a\ triangle\ ABC\ if\ two\ vertives\ are\ A(-3,1)\ and\ B(0,2)\ while\ midpoint\ of\ BC\ is\ at\ point\ D\left(\frac{3}{2},\ -\frac{1}{2}\right)$  

• (3,-3)

• (-3,3)

• (3,3)

• (0,3)