Question

Q6. Find a relation between x and y such that the point (x , y) is equidistant from the points (5, 1) and (4, 7).​

Answer 1

Answer:

(x+5y)=-39/2

Step-by-step explanation:

PA=PB

PA^2=PB^2

(X-5)^2+(Y-1)^2=(X-4)^2+(Y-7)^2

X^2-10X+25+Y^2-2Y+1=x^2-8x+16+y^2-14y+49

-10X-2Y+26=-8X-14Y+65

-10X+8X-2Y+14Y=65-26

-2X+10Y=39

-2(X+5Y)=39

x+5y=-39/2