Question

If x-y=2 then point (x,y) is equidistant from (7,1) and _______

Answer 1

Point (x,y) is equidistant from (7,1) and (3, 5)

Step-by-step explanation:

Consider the point (a, b) such that,

(x,y) is equidistant from (7,1) and (a, b),

That is, (x, y) is midpoint of line joining (7, 1) and (a, b)

$\implies (\frac{7+a}{2}, \frac{1+b}{2})=(x, y)$

On comparing the coordinates,

$\frac{7+a}{2}=x, \frac{1+b}{2}=y$

7 + a = 2x, 1 + b = 2y

⇒ 2x - 2y = 7 + a - (1+b)  = 7 + a - 1 - b = 6 + a - b

⇒ 2(x-y) = 6 + a - b

We have,

x - y = 2,

⇒ 2(2) = 6 + a - b

⇒ 4 = 6 + a - b

⇒ a - b = -2    .....(1)

Now, line joining (a, b) and (7, 1) will be perpendicular to x - y =2,

∵ x - y = 2 ⇒ y = x - 2 ⇒ slope of x - y = 2 is 1  ( comparing with line y = mx + c, where m is the slope )

Slope of line joining (a, b) and (7, 1) = $\frac{1-b}{7-a}$

By property of perpendicular lines,

Slope of 'x - y = 2' × slope of line joining (a, b) and (7, 1) = - 1

$1\times \frac{1-b}{7-a}=-1$

$\frac{1-b}{7-a}=-1$

$1-b=-(7-a)$

$1-b=a-7$

$\implies a + b = 8$     ....(2)

Equation (1) + equation (2)

2a = 6

a = 3

From equation (1),

3 - b = -2

-b = -2 - 3

-b = -5

⇒ b = 5

Hence, point (x,y) is equidistant from (7,1) and (3, 5)

#Learn more :

Find the equation of the line through (-2,3) with slope -4

https://brainly.in/question/11352276

Find midpoint :

https://brainly.in/question/12134707

Answer 2

Answer: x= 3, y=5

Step-by-step explanation: