Question

If A=(6,-4)B=(-2,2),then find the co-ordinates of "P" such that PA=PB and A,B,P are collinear​

Answer 1

Answer:

2 is the answer of these questions

Answer 2

The given

PA=PB

distance of PA = distance of PB

let the point of P = (x,y)

and name of points

X1 =6. , Y1= -4

X2 = X , Y2 = Y

X3 = -2 ,Y3= 2

$\sqrt{ {(x2 - x1)}^{2} + {(y2 - y1)}^{2} } = \sqrt{ {(x3 - x2)}^{2} + {(y3 - y2)}^{2} } \\ {(x2 - x1)}^{2} + {(y2 - y1)}^{2} = {(x3 - x2)}^{2} + {(y3 - y2)}^{2} \\ {(x - 6)}^{2} + {(y + 4)}^{2} = {( - 2 - x)}^{2} + {(2 - y)}^{2} \\ {x }^{2} + {6}^{2} - 2 \times 6 \times x + {y}^{2} + {4}^{2} + 2 \times 4 \times y = {2}^{2} + {x}^{2} + 2 \times 2 \times x + {2}^{2} + {y}^{2} - 2 \times y \times 2 \\ 36 - 12x + 16 + 8y - 4 - 4x - 4 + 4y = {x}^{2} - {x}^{2} + {y}^{2} - {y}^{2} \\ - 16x + 12y + 36 + 8 = 0 \\ 16x - 12y = 44 \\ 4(4x - 3y) = 44 \\ 4x - 3y = \frac{44}{4} \\ 4x - 3y = 11 \: \: equation \: 1$

second type A,B,P are collinear

solve

$x1(y2 - y3) + x2(y3 - y1) + x3(y2 - y1) = 0 \\ 6(y - 2) + x(2 - 4) + ( - 2)(y - 4) = 0 \\ 6y - 12 - 2x - 2y + 8 = 0 \\ 4y - 2x - 4 = 0 \\ 4y - 2x = 4 \\ 2(2y - x) = 4 \\ 2y - x = 2 \: \: equation \: 2$

do multiple 4 in equation 2

so equation = 8y-4x = 8 equation 3

Add the equation 1 and 3

$4x - 3y = 11 \\ - 4x + 8y = 8 \\ add \: the \: equation \\ - 3y + 8y = 11 + 8 \\ 5y = 19 \\ y = \frac{19}{5} put \: on \: eqution \: 2$

$2y - x = 2 \\ 2 \times \frac{19}{5} - x = 2 \\ \frac{38}{5} - x = 2 \\ \frac{38}{5} - 2 = x \\ \frac{38 - 10}{5} = x \\ x = \frac{28}{5}$

X= 28/5

and

Y = 19/5

points Of P = (28/5,19/5) answer