Question

solve karo aata toh yaar ​

Answer 1

Answer:

The renquired points which are

equidistant from the given points and

lie on x-axis are (5,0) and (17,0)

Step-by-step explanation:

Given :-

Given point A(11,-8) and Distance is 10 units

To find:-

Find the points on the x-axis which is at equidistant from A

Solution:-

Let the required point which is on the x-axis be

(x,0)

since the equation of x-axis is y=0

And given point is A(11,-8)

Using formula:-

If(x1,y1) and (x2,y2) are the points then the distance between them is

√[(x2-x1)²+(y2-y1)²] units

now

we have (x1,y1)=(11,-8)=>x1=11;y1=-8

and (x2,y2)=(x,0)=>x2=x;y2=0

Distance between the given points=10 units

=>√[(x-11)²+{(0-(-8)}²=10

=>√[(x-11)²+8²]=10

=>√[x²-2(x)(11)+(11)²+8²]=10

=>√[x²-22x+121+64]=10

=>√[x²-22x+185]=10

on squaring both sides

=>{√[x²-22x+185]}²=10²

=>x²-22x+185=100

=>x²-22x+185-100=0

=>x²-22x+85=0

=>x²-5x-17x+85=0

=>x(x-5)-17(x-5)=0

=>(x-5)(x-17)=0

=>x-5=0 or x-17=0

x=5 and x=17

The Points (5,0) and (17,0)

Answer:-

The renquired points which are equidistant from the given points and lie on x-axis are (5,0) and (17,0)

Answer 2

Answer:

Given:

distance = 10

one of point is given A (11 , -8)

To Find:

another point

Proof :

distance = root of (x2-x1)^2 + (y2-y1)^2

=> assume that another point as B (x,0)

=> 10 = root of (x-11)^2 + 64

here root transfer to lhs side

=> 100 = x^2 + 121 - 22x + 64

=> 36 = x^2 + 121 - 22x

=> x^2 - 22x + 85 = 0

=> x^2 - 17x - 5x + 85 = 0

=> x (x - 17) - 5 (x - 17) = 0

=> (x - 17) ( x - 5) = 0

=> x = 17 , 5

The second point B (17 , 0) or (5,0)