Question

Find the point on x axis which are at a distance of 2under root 5 unit from the point (7,-4).how many such point are there?

Answer 1

HELLO DEAR,


LET THE POINT ON X-AXIS BE (x,0)

AND GIVEN THAT:-AB=2√5


$2 \sqrt{5} = \sqrt{( {7 - x})^{2} + ( { - 4 - 0})^{2} } \\ = > ( {2 \sqrt{5} })^{2} = ( {7 - x}) ^{2} + 16 \\ = > 20 = 49 + {x}^{2} - 14x + 16 \\ = > {x}^{2} - 14x + 45 = 0 \\ = > {x}^{2} - 9x - 5x + 45 = 0 \\ = > x( x - 9) - 5(x - 9) \\ = > (x - 5)(x - 9) = 0 \\ = > x = 9 \: \: and \: \: x = 5$
there are two points i.e., (5, 0) and (9, 0).


I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

The answer to this question is
2√​5​​​=√​(7−x)​2​​+(−4−0)​2​​​​​​=>(2√​5​​​)​2​​=(7−x)​2​​+16​=>20=49+x​2​​−14x+16​=>x​2​​−14x+45=0​=>x​2​​−9x−5x+45=0​=>x(x−9)−5(x−9)​=>(x−5)(x−9)=0​=>x=9andx=5​​ 
there are two points i.e., (5, 0) and (9, 0).