ICSE Board Question Based on Section Formula Chapter of M.L Aggarwal for class10

In this question given a point is reflected to another point in the x-axis.Find the solution of the question

This is the Question Number 07, Exercise 11 of M.L Aggarwal.

# Point P (3, – 5) is reflected to P’ in the x- axis. Also P on reflection in the y-axis is mapped as P”. (i) Find the co-ordinates of P’ and P”. (ii) Compute the distance P’ P”. (iii) Find the middle point of the line segment P’ P”. (iv) On which co-ordinate axis does the middle point of the line segment P P” lie ?

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(i) The image of P(3,-5) when reflected in X-axis will be (3,5).When you reflect a point across the X-axis, the x-coordinate remains the same,

but the y-coordinate is transformed into its opposite (its sign is changed).

Co-ordinates of P’ = (3,5)Image of P(3,-5) when reflected in Y axis will be (-3,-5).When you reflect a point across the Y-axis, the y-coordinate remains the same,

but the x-coordinate is transformed into its opposite (its sign is changed)

Co-ordinates of P’’ = (-3,-5)(ii)Let P’(x

_{1}, y_{1}) and P’’(x_{2}, y_{2}) be the given pointsBy distance formula d(P’,P’’) = √[(x

_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]Co-ordinates of P’ = (3,5)Co-ordinates of P’’ = (-3,-5)Here x

_{1}= 3, y_{1}= 5 , x_{2}= -3, y_{2}= -5d(P’,P’’) = √[(-3-3)

^{2}+(-5-5)^{2}]= √[(-6)

^{2}+(-10)^{2}]= √(36+100)

= √136

= √(4×34)

= 2√34

Hence the distance between P’ and P’’ is 2√34 units.

(iii)

Co-ordinates of P’ = (3,5)Co-ordinates of P’’ = (-3,-5)Here x

_{1}= 3, y_{1}= 5 , x_{2}= -3, y_{2}= -5Let Q(x,y) be the midpoint of P’P’’

By midpoint formula,

x = (x

_{1}+x_{2})/2y = (y

_{1}+y_{2})/2x = (3+-3)/2 = 0/2 = 0

y = (5+-5)/2 = 0/2 = 0

Hence the co-ordinate of midpoint of P’P’’ is (0,0) .

(iv)

Co-ordinates of P = (3,-5)Co-ordinates of P’’ = (-3,-5)Here x

_{1}= 3, y_{1}= -5 , x_{2}= -3, y_{2}= -5Let R(x,y) be the midpoint of PP’’

By midpoint formula,

x = (x

_{1}+x_{2})/2y = (y

_{1}+y_{2})/2x = (3+-3)/2 = 0/2 = 0

y = (-5+-5)/2 = -10/2 = -5

So the co-ordinate of midpoint of PP’’ is (0,-5) .

Here x co-ordinate is zero.

Hence the point lies on Y-axis.