Question

An exhibition is being organised in the school premises to highlight 'Swatchchh
Bharat Abhiyan' and 'Say no to plastic'. Material concerning these is displayed at
the positions represented by the points A(-4,0) and B(4,0). A spot light is placed
at a point represented by P(a, b) such that A, B and P are equidistant from each
other.
(1) Help me in finding the position of point P by finding the values of a and b.
(ii) The distance of point P from line segment AB.​

Answer 1

(i)Therefore the coordinate of P is (0, $2\sqrt{14}$)

(ii)Therefore the distance of point (0,$2\sqrt{14}$) and from the line segment AB is

= 4 units

Step-by-step explanation:

Given, points are A(-4,0) and B(4,0).

The distance between A(-4,0) and B(4,0) is$\sqrt{(4+4)^2+(0-0)^2}$= 8

The distance between A(-4,0) and P(a,b) is $\sqrt{(a+4)^2+(b-0)^2}$

The distance between A(4,0) and P(a,b) is $\sqrt{(a-4)^2+(b-0)^2}$

According to the problem,

$\sqrt{(a+4)^2+(b-0)^2}$=$\sqrt{(a-4)^2+(b-0)^2}$

⇒(a+4)²+b²=(a-4)²+b²

⇒a²+8a+16 +b²=a²-8a+16+b²

⇒8a =0

⇒a=0

and $\sqrt{(a+4)^2+(b-0)^2}$ =8

⇒(a+4)²+b²=64

⇒16+b²=64

⇒b²=56

⇒b=$2\sqrt{14}$

Therefore the coordinate of P is (0, $2\sqrt{14}$)

The equation of line passes through A and B is

(0-0)(y-0)=(4+4)(x-4)                     $[(y_2-y_1)(y-y_2)=(x_2-x_1)(x-x_2)]$

⇒x=4

Therefore the distance of point (0,$2\sqrt{14}$) and from the line segment AB is

$=|\frac{0-4}{\sqrt{1^2} } |$ = 4 units