Math, asked by sonusharma45, 5 months ago

9 number koi solve kar do yaar​

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Answers

Answered by tennetiraj86
4

Step-by-step explanation:

The renquired points which are

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

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

Answer:-

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

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