For the 1st one……

use the distance formula……its the shortest way…..

Let the coordinates of that point be (x,y)

Now

square root of……..{ [x-(-5)] to the power 2……plus (y-5) to the power 2 } will be equal to

square root of…….{ [x-(-2)] to the power 2…..plus (y-2) to the power 2 }……..equate the two…..and get an equation in terms of x and y…….that is the locus….!!!!

For the 2nd….

Again….

square root of……..{ (x-1) to the power 2……plus [y-(-7)] to the power 2 } will be equal to 5.

Again u get an equation…..which is the locus…!!!!

provided that ABC is a triangle. Whoose verices A(2,5) , B(4,-3) , C(-4 , a million) Meadian is a line phase from a vertex of a triangle to the midpoint of the alternative component right here you prefer discover the equation of meadian from the vertex B to the alternative component AC midpoint of the component AC = ((2+ (-4))/2 , (5+a million)/2) = (-2/2 , 6/2) = (-a million,3) { interior the above i used the formulation for midpoint of linesegment = ( ( x1+x2)/2 , (y1+y2)/2 ) } midpoint of the component AC, D = (-a million , 3) {call the midpoint as D} slope of the meadian BD = (-3-3) / (4-(-a million)) = -6/5 {interior the above i used the formulation for the slope of line = (y2-y1)/(x2-x1) } now the equation of the meadian BD :- (y- (-3) ) = -6/5 (x – 4) {interior the above i used the line equation slope – component formulation (y-y1) = m (x-x1) } equation of the meadian :- 5(y+3) = -6 (x – 4) 5y + 15 = -6x +24 6x + 5y +15 – 24 =0 6x +5y -9 =0 so the equation of the meadian type B to AC is 6x +5y -9 =0

hey your question is easy but i cant show the solution here. it is quite hard to type the equations including the step by step solution. if u want u can email me at [email protected]

ill show you how to do it? ok?