Question

Find the coordinates of the points which lies on the line 2x-5y=7,and is equidistant from the point A(-2,3) and B(3,4)

Answer 1

Answer:

(37/27, - 23/27 )

Step-by-step explanation:

Let that point be ( a, b ).

As ( a, b ) lies on 2x - 5y = 7,

= > 2(a) - 5(b) = 7

= > 2a - 5b = 7

= > 5a = 5/2 ( 7 + 5b ) ... (1)

As it is equidistant from the given points.

= > distance b/w ( -2, 3 ) and ( a, b ) = distance b/w ( 3, 4 ) and ( a, b )

Using distance formula,

= > √{ ( - 2 - a )² + ( 3 - b )² } = √{ ( 3 - a )² + ( 4 - b )² }

= > ( - 2 - a )² + ( 3 - b )² = ( 3 - a )² + ( 4 - b )^2

= > 4 + a² + 4a + 9 + b² - 6b = 9 + a² - 6a + 16 + b² - 8b

= > 10a + 2b = 12

= > 5a + b = 6

= > (5/2) ( 5b + 7 ) + b = 6 { 5a = (5/2) ( 5b + 7 ) }

= > 25b + 35 + 2b = 12

= > 27b = - 23

= > b = - 23/27

Hence, a = (1/2) ( 5(-23/27) + 7 ) = 37/27

Hence the required point is ( 37/27, - 23/27 ).