Question

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

Answer 1

Answer:

• it is the difference between x and y (7,1) is x=1,y=7 and(-1,-4)is xaxis is 4 taxis is 1 , because (-1,-7 ) is the direction of opposite side.

Answer 2

❏ Question:-

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

❏ Solution:-

✏ Distance between the points (x,y) and (7,1) is,

$\sf\implies D_1= \sqrt{(x-7)^2+(y-1)^2}$ units

✏ Distance between the points (x,y) and (-1,-4) is,

$\sf\implies D_2= \sqrt{[x-(-1)]^2+[y-(-4)]^2}$ units

Now, according to the question the point (x,y) is Equidistant from the points (7,1) & (-1,-4),

$\sf\therefore D_1=D_2$

$\sf\implies\sqrt{(x-7)^2+(y-1)^2}=\sqrt{[x-(-1)]^2+[y-(-4)]^2}$

$\sf\implies (x-7)^2+(y-1)^2=(x+1)^2+(y+4)^2$

$\sf\implies x^2-2\times x\times7+7^2+y^2-2\times y\times1+1^2= x^2+2\times x\times1+1^2+y^2+2\times y\times4+4^2$

$\sf\implies \cancel{x^2}-14x+49+\cancel{y^2}-2y+1= \cancel{x^2}+2x+1+\cancel{y^2}+8y+16$

$\sf\implies -14x+49-2y+1= 2x+1+8y+16$

$\sf\implies 50-14x-2y= 17+2x+8y$

$\sf\implies 50-17=2x+8y+14x+2y$

$\sf\implies 33=16x+10y$

$\sf\implies \boxed{\large{\red{16x+10y=33}}}$