Question

A (-5,2) and B (4,1) find the equation of the locus of point P, which is equidistant from A and B .....Help me guys​

Answer 1

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• A($\sf x_2, y_2$) = (-5, 2)
• B($\sf x_3, y_3$)= (4, 1)
• Point P is equidistant from A and B

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• Equation of locus of point P.

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• Let the locus of point P be (h,k).
• Also let P = ($\sf x_1, y_1$) = (h, k)

As, it is given that P is equidistant from A an B.

Therefore, Length of AP = Length of BP

Using Distance formula on AP and BP

$\boxed{ \sf \: AP = \sqrt{ {(x_1 - x_2 )}^{2} \: + \: {(y_1 - y_2 )}^{2} } }$

$\boxed{ \sf \: BP = \sqrt{ {(x_1 - x_3 )}^{2} \: + \: {(y_1 - y_3)}^{2} }}$

$\therefore \sf{As \: \: AP = BP }$

$\implies \sf \: \sqrt{ {(x_1 - x_2 )}^{2} \: + \: {(y_1 - y_2 )}^{2} } = \sqrt{ {(x_1 - x_3 )}^{2} \: + \: {(y_1 - y_3)}^{2} }$

Squaring both sides we get,

$\implies \sf \: ({ \sqrt{ {(x_1 - x_2 )}^{2} \: + \: {(y_1 - y_2 )}^{2} }})^{2}= ( {\sqrt{ {(x_1 - x_3 )}^{2} \: + \: {(y_1 - y_3)}^{2} }})^{2}$

$\implies \sf \: {(x_1 - x_2 )}^{2} \: + {(y_1 - y_2 )}^{2} = {(x_1 - x_3 )}^{2} \: + \: {(y_1 - y_3)}^{2}$

$\implies \sf \: {(h - ( - 1) )}^{2} \: + {(k - 2)}^{2} = {(h - 4 )}^{2} \: + \: {(k - 1)}^{2}$

$\implies \sf \: {(h + 5)}^{2} \: + {(k - 2)}^{2} = {(h - 4 )}^{2} \: + \: {(k - 1)}^{2}$

$\sf{ \implies \: {h}^{2} + 25 + 10h \: + \: {k}^{2} + 4 - 4k \: } = {h}^{2} + 16 - 8h + {k}^{2} + 1 - 2k$

$\sf{ \implies \: \cancel{h}^{2} + 29 + 10h \: + \: \cancel{k}^{2} - 4k \: = \cancel {h}^{2} + 17 - 8h + \cancel{k}^{2} - 2k}$

$\sf{ \implies \: 10h \: - 4k \: + 29 = - 8h - 2k + 17}$

$\sf{ \implies \: (10h + 8h)\: - (4k - 2k) \: +( 29 - 17) = 0}$

$\sf{ \implies \: 18h\: - 6k\: + 12= 0}$

$\sf{ \implies \: 3h\: - k\: + 2= 0}$

[Putting, h --> x, k --> y]

$\boxed{ \sf{ \therefore \: 3x\: - y\: + 2= 0}}$

Hence, the equation of locus of point P is 3x - y + 2 = 0

#answerwithquality # BAL

Answer 2

Answer:

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