Question 1

Point on the  x -axis which is equidistant from  (2, -5)  and  (-2, 9)  is:

Accepted Answer

Since point on  x- axis, then coordinate of the point is  \left(x, 0ight ) . According to the question this point  \left(x, 0ight )  is equidistant from the points  \left(2, -5ight)  and  \left(-2, 9ight) .That is, distance from   \left(x, 0ight )  and  \left(2, -5ight)   =  distance from  \left(x, 0ight )  and  \left(-2, 9ight) . \sqrt{\left( 2-x ight)^2+\left( -5-0 ight)^2}=\sqrt{\left( -2-x ight)^2+\left( 9-0 ight)^2}\ \Rightarrow \left( 2-x ight)^2+\left( -5 ight)^2=\left( -2-x ight)^2+\left( 9 ight)^2\ \Rightarrow 4-4x+x^2 +25=4 +4x+4+81 \ \Rightarrow 8x=25-81\ \Rightarrow 8x=-56\ \Rightarrow x =-7  So, point is  \left(-7,0 ight) .

Question 2

Find a relation between  x  and  y  such that the point  (x, y)  is equidistant from the point  (3, 6)  and  ( -3, 4) .

Accepted Answer

Let the point  P(x,y)  is equidistant from the points  A(3,6) and  B(-3,4) 	herefore PA=PB By using the distance formula =\sqrt{ { \left( {x_2}-{ { x_1 } } ight)  }^{ 2}+{ \left( {y_2 }-{ y_1  } ight)  }^{ 2 } }      we have \Rightarrow \sqrt{(x-3)^2+(y-6)^2}=\sqrt{(x+3)^2+(y-4)^2} \Rightarrow (x-3)^2+(y-6)^2=(x+3)^2+(y-4)^2 \Rightarrow x^2-6x+9+y^2-12y+36=x^2+6x+9+y^2-8y+16 \Rightarrow -6x-6x-12y+8y+36-16=0 \Rightarrow -12x-4y+20=0 \Rightarrow 3x+y-5=0 Hence this is a relation between  x  and  y .

Question 3

The perimeter of a triangle with vertices  (0, 4) (0, 0)  and  (3, 0)  is:

Accepted Answer

Let  ABC  be a triangle having  A(0,4); B(0,0); C(3,0)  Using distance formula, we get  AB=\sqrt { { (0-0) }^{ 2 }+{ (0-4) }^{ 2 } }  =\sqrt { 16 }  =4\ BC=\sqrt { { (3-0) }^{ 2 }+{ (0-0) }^{ 2 } }  =\sqrt { 9 }  =3\ CA=\sqrt { { (0-3) }^{ 2 }+{ (4-0) }^{ 2 } } =\sqrt { 25 }  =5 Perimeter of  	riangle ABC=AB+BC+CA=4+3+5=12

Question 4

If the distance between the points  (4, p)  and  (1, 0)  is  5 , then the value of  p  is:

Accepted Answer

The given points are  (1,0)  and   (4,p)    d=\sqrt { { \left( { x }_{ 2 }-{ x }_{ 1 } \right)  }^{ 2 }+{ \left( { y }_{ 2 }-{ y }_{ 1 } \right)  }^{ 2 } } \Rightarrow 5 =\sqrt{(4-1)^2+p^2} \Rightarrow 25 =(4-1)^2+p^2 \Rightarrow 25-9=p^2 \Rightarrow p^2=16 \Rightarrow p=+4,-4

Question 5

Find a relation between x and y such that the point (x,y) is equidistant from the points (7,1) and (3,5).

Accepted Answer

Let P(x,y) be equidistant from the points A(7,1) and B(3,5). AP=BP AP^2=BP^2 (x-7)^2+(y-1)^2=(x-3)^2+(y-5)^2 x^2-14x+49+y^2-2y+1=x^2-6x+9+y^2-10y+25 x-y=2

Question 6

Find the coordinates of the foot of perpendicular from the point  (-1, 3)  to the line  3x -  4y -  16\displaystyle = 0

Accepted Answer

Let  (a,b)  be the coordinates of the foot of the perpendicular from 
the point  (-1,3)  to the line  3x  -4y -  16\displaystyle = 0 Slope of the line joining  (-1,3)  and  (a,b) , is  \displaystyle m_{1}= \frac{b-3}{a+1} Slope of the line  3x  -4y -  16\displaystyle = 0  or  \displaystyle y= \frac{3}{4}x-4,m_{2}= \frac{3}{4} Since these two lines are perpendicular, \displaystyle m_{1}m_{2}= -1 \displaystyle 	herefore \left \{ \frac{b-3}{a+1} ight \}\cdot \left \{ \frac{3}{4} ight \}= -1 \displaystyle \Rightarrow \frac{3b-9}{4a+4}= -1 \displaystyle \Rightarrow 3b-9= 4a-4 \displaystyle \Rightarrow 4a+3b= 5 ...(1)Point(a,b) lies on line  \displaystyle 3x-4y= 16 \displaystyle 	herefore 3a-4b= 16 ....(2)On solving equation (1) and (2) we obtain \displaystyle a= \frac{68}{25}  and  b= -\cfrac{49}{25} Thus,
 the required coordinates of the foot of the perpendicular are 
 \displaystyle \left \{ \frac{68}{25},\frac{49}{25} ight \}

Question 7

If  A(-2, 1), B(a, 0), C(4, b)  and  D(1, 2)  are the vertices of a parallelogram ABCD, find the values of a and b. Hence find the lengths of its sides.

Accepted Answer

Given sides of parallelogram  A(-2 , 1) ,B(a,0) ,  C(4 ,b)  , D(1 , 2) We know that diagonals of Parallelogram bisect each other. Mid-point let say   O  of diagonal  AC  is given by  x=\left ( \dfrac{x_1+x_2}{2}\right ) and \ y=\left ( \dfrac{y_1+y_2}{2}\right ) O    \left(\dfrac{-2+4}{2}  , \dfrac{1+b}{2}\right)  ..................(1)Mid-point let say   P  of diagonal  BD  is given by  P    \left(\dfrac{a+1}{2}  , \dfrac{0+2}{2}\right)  ..................(2)Points  O  and  P  are same Equating the corresponding co-ordinates of both midpoints, we get \dfrac{-2+4}{2}  = \dfrac{a+1}{2}   \Rightarrow a=1 and \dfrac{1+b}{2}  = \dfrac{0+2}{2}   \Rightarrow b=1 Now the Given co-ordinates of the parallelogram are written as  A(-2 , 1) ,B(1,0) ,  C(4 ,1)  , D(1 , 2) By distance formula,  \sqrt{ { \left( {x_2}-{ { x_1 } } \right)  }^{ 2}+{ \left( {y_2 }-{ y_1  } \right)  }^{ 2 } }   we can find the length of each side  AB = \sqrt{(-2-1)^2+(1-0)^2} AB = \sqrt{(3)^2+(1)^2}  = \sqrt{10} AB=CD    ...............(pair of opposite sides of the parallelogram are parallel and equal) BC = \sqrt{(4-1)^2+(1-0)^2} BC = \sqrt{(3)^2+(1)^2}  = \sqrt{10} BC=AD    ...............(pair of opposite sides of the parallelogram are parallel and equal   ) \Rightarrow AB=BC=CD=AD = \sqrt{10} \Rightarrow ABCD   is a   Rhombus

Question 8

Find the distance of the point  P(-6,8)  from the origin.

Accepted Answer

The distance of the point  P(-6,8)  from the origin is  \sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10  units.

Question 9

If  (1, 2), (4, y), (x, 6)  and  (3, 5)  are the vertices of a parallelogram taken in order, find  x  and  y .

Accepted Answer

Let  A(1,2) ,  B(4,y) , C(x,6)  and  D(3,5)  are the vertices of a parallelogram  ABCD .AC  and  BD  are the diagonals . O  is the midpoint of  AC  and  BD. If  O  is the mid-point of  AC  ,then the coordinates of  O  are  =\left(\dfrac{1+x}{2},\dfrac{2+6}{2} ight) = \left(\dfrac{x+1}{2},4 ight) If  O  is the mid-point of BD then coordinates of O are  \left(\dfrac{4+3}{2},\dfrac{5+y}{2}ight) = \left(\dfrac{7}{2},\dfrac{5+y}{2} ight) Since both coordinates are of the same point  O 	herefore \dfrac{1+x}{2}=\dfrac{7}{2} \Rightarrow 1+x=7 \Rightarrow x=7-1=6 	herefore \dfrac{5+y}{2}=4 \Rightarrow 5+y=8 \Rightarrow y=8-5=3 Hence   x=6  and  y=3 .

Question 10

Find the distance of a point  P(x, y)  from the origin.

Accepted Answer

Let  O  be the origin and  Q  be the foot of the perpendicular dropped from  P  onto the  x  axis. So  \Delta OPQ  is right-angled at  Q .By definition of coordinates: OQ=x  coordinate of  P=  distance of  P  from  y  axis  = |x| Similarly,  QP = |y| .Thus, by using Pythagoras theorem on  \Delta OPQ , we get  OP = \sqrt{OQ^2 + QP^2} = \sqrt{x^2 + y^2}  Thus the distance of the point  P(x, y)  from the origin  O(0, 0)  is  \sqrt{x^2+y^2}