Question

Find a point on the y axis which is equidistant from A (2,-5) and (-4,3) . ​

Answer 1

TO FIND :–

• A point on the y - axis which is equidistant from A (2,-5) and B(-4,3) .

SOLUTION :–

• Let the point on y - axis is P(0 , y).

• We know that Distance between two points $\: \bf (x_1 , y_1 ) \:\: and \:\: (x_2 , y_2 ) \:$ is –

$\\ \large \implies{ \boxed{ \bf Distance = \sqrt{(x_2 -x_1)^{2} +(y_2 -y_1)^{2}}}}$

• Let's find Distance AP –

$\\ \large \implies \bf AP = \sqrt{(0 - 2)^{2} +(y - ( - 5))^{2}}$

$\\ \large \implies \bf AP = \sqrt{(- 2)^{2} +(y + 5)^{2}}$

$\\ \large \implies \bf AP = \sqrt{4+(y + 5)^{2}} \: \: \:\: - - -eq.(1)$

• Now Let's find BP –

$\\ \large \implies \bf BP = \sqrt{(0 - (-4))^{2} +(y -3)^{2}}$

$\\ \large \implies \bf BP = \sqrt{(4)^{2} +(y - 3)^{2}}$

$\\ \large \implies \bf BP = \sqrt{16+(y-3)^{2}} \: \: \:\: - - -eq.(2)$

• According to the question –

$\\ \large \implies \bf AP =BP$

• Using eq.(1) & eq.(2) –

$\\ \large \implies \bf \sqrt{4+(y + 5)^{2}} =\sqrt{16+(y-3)^{2}}$

• Square on both sides –

$\\ \large \implies \bf 4+(y + 5)^{2} =16+(y-3)^{2}$

$\\ \large \implies \bf (y + 5)^{2} =12+(y-3)^{2}$

$\\ \large \implies \bf (y + 5)^{2} - (y-3)^{2} =12$

$\\ \large \implies \bf (y+5 +y-3) \{y + 5 - (y-3) \} =12$

$\\ \large \implies \bf (2y + 2) \{y + 5 - y + 3 \} =12$

$\\ \large \implies \bf (2y + 2)(8)=12$

$\\ \large \implies \bf (2y + 2)(2)=3$

$\\ \large \implies \bf (y+1)(4)=3$

$\\ \large \implies \bf y= \dfrac{3}{4} - 1$

$\\ \large \implies{ \boxed{ \bf y= - \dfrac{1}{4}}}$

▪︎ Hence , The point is (0 , - ¼).

Answer 2

$\large \rm \red{ \underline{ \underline {Given:-}}}$

• Two points A(2, -5) and B(-4, 3)

$\large \rm \blue{ \underline{ \underline {To\:Find:-}}}$

• A point on the y axis which is equidistant from A (2,-5) and B(-4,3) .

$\large \rm \green{ \underline{ \underline {To\:Find:-}}}$

$\rm the \: distance(d) \: between \: two \: points \: ( x_{1} ,y_{1}) \: and \: ( x_{2} ,y_{2}) \: is \: given \: by \: the \: formula:- \:$

$\rm d = \sqrt{ {( x_{2} - x_{1}) }^{2} + {( y_{2} - y_{1}) }^{2} }$

As we have have to find the point on the y-axis equidistant from the points A(2, -5) and B(-4, 3)

Let us denote that point as C(x , y)

As the point lies on the y-axis, so x = 0

Let us find the distance between A to C and B to C

$\rm AC = \sqrt{ {( 2 - x) }^{2} + {( - 5 - y) }^{2} }$

$\rm AC = \sqrt{ {( 2 - 0) }^{2} + {( - 5 - y) }^{2} }$

$\rm AC = \sqrt{ {( 2 ) }^{2} + {( - 5 - y) }^{2} }$

$\rm AC = \sqrt{ 4 + {( - 5 - y) }^{2} }$

$\rm BC= \sqrt{ {( - 4 - x)}^{2} + {( 3 - y) }^{2} }$

$\rm BC= \sqrt{ {( - 4 - 0)}^{2} + {( 3 - y) }^{2} }$

$\rm BC= \sqrt{ {( - 4 )}^{2} + {( 3 - y) }^{2} }$

$\rm BC= \sqrt{ 16 + {( 3 - y) }^{2} }$

The point equidistant from y-axis

$\rm \rightarrow \sqrt{ 4 + {( - 5 - y) }^{2} } = \sqrt{ 16 + {( 3 - y) }^{2} }$

S.O.B.S

$\rm \rightarrow { 4 + {( - 5 - y) }^{2} } = { 16 + {( 3 - y) }^{2} }$

$\rm \rightarrow { 4 + {( { - 5)}^{2}+ (- y) }^{2} + 2( - 5)( - y) } = { 16 + {( {3)}^{2}+ ( - y) }^{2} + 2(3)( - y) }$

$\rm \rightarrow { 4 + 25 + {y}^{2} +10y} = { 16 + { {9} + y }^{2} -6 y }$

$\rm \rightarrow { 29 +10y} = { 25 -6 y }$

$\rm \rightarrow { 10y + 6y} = { 25 - 29 }$

$\rm \rightarrow { 16y} = { - 4 }$

$\rm \rightarrow { y} = { \dfrac{ - 4}{16} }$

$\rm \rightarrow { y} = { \dfrac{ - 1}{4} }$

Hence the point on the y-axis is (0, -¼)