Question

10. Find the point on x-axis which is equidistant from (7,6) and (-3, 4).​

Answer 1

Step-by-step explanation:

• the point P (3,0)
• I think it must help you

Answer 2

GIVEN :-

• ( X1 , Y1 ) = ( 7 , 6 )

• ( X2 , Y2) = ( -3 , 4)

• ( X3 , Y3) = ( X , 0 )

• x-axis which is equidistant from (7,6) and (-3, 4).

TO FIND :-

• value of ( X3 , Y3 )

SOLUTION :-

FOR X1 AND Y1

we know that they are equidistant to x axis hence ,

$\implies \rm{ \bf{distance (x _{1} </p><p>,y _{1}) =distance (x _{3} ,y _{3})}}$

now we know the distance formula -

$\implies \boxed{\rm{ \sqrt{ (x _{3} - x_{1}) {}^{2} + (y _{3} - y _{1}) {}^{2} \: \:}}}$

$\implies \rm{ \sqrt{ (x - 7) {}^{2} + (0 - 6) {}^{2} \: \:}}$

$\implies \rm{ \sqrt{ x {}^{2} + (7) {}^{2} - 14x + 36 \: \:}}$

$\implies \rm{ \sqrt{ x {}^{2} + 49 - 14x + 36 \: \:}}$

$\implies \rm{ \sqrt{ x {}^{2} - 14x +85\: } } \: \: \: \: \: \: \: \: (1)$

FOR X2 AND Y2

we know that they are equidistant to x axis hence ,

$\implies \rm{ \bf{distance(x _{2} </p><p>,y _{2}) =distance (x _{3} ,y _{3})}}$

$\implies \boxed{\rm{ \sqrt{ (x _{3} - x_{2}) {}^{2} + (y _{3} - y _{2}) {}^{2} \: \:}}}$

$\implies \rm{ \sqrt{ (x + 3) {}^{2} + (0 - 4) {}^{2} \: \:}}$

$\implies \rm{ \sqrt{ {x}^{2} +( 3 ){}^{2} + 6x + 16\: \:}}$

$\implies \rm{ \sqrt{ {x}^{2} +9+ 6x + 16\: \:}}$

$\implies \rm{ \sqrt{ {x}^{2} + 6x + 25}} \: \: \: \: \: \: (2)$

now we know that eq 1 = eq 2 as both are same distance so

$\implies \rm{ \sqrt{ {x}^{2} - 14x + 85 \: } = \sqrt{ {x}^{2} + 6x + 25 \:}}$

squaring both sides

$\implies \rm{ (\sqrt{ {x}^{2} - 14x + 85 \: } ) {}^{2} =( \sqrt{ {x}^{2} + 6x + 25 \:}) {}^{2} }$

$\implies \rm{ {x}^{2} - 14x + 85 \: = {x}^{2} + 6x + 25 \:}$

$\implies \rm{ - 14x + 85 \: = 6x + 25 \:}$

$\implies \rm{ 20x = 60 \:}$

$\implies \rm{ x = 3}$

thus point is :-

$\implies \boxed{ \boxed{\rm{ (x _{3}</p><p></p><p>,</p><p> y_{3} ) = (3 </p><p></p><p>,</p><p> 0)}}}$