Question

If x minus y equal to 2 then point (x,y) is equidistant from (7, 1) and

Answer 1

Given unknown point equidistant from the point (x, y) is (3, 5).

Step-by-step explanation:

From the figure attached points A(7, 1) and B(a, b) are equidistant from the given line x - y = 2

Therefore, O is the center of line segment AB.

Coordinates of O,

x = $\frac{(a+7)}{2}$

and y = $\frac{(b+1)}{2}$

Now we plug in these values in the given equation x - y = 2

$\frac{(a+7)}{2}-\frac{(b+1)}{2}=2$

a + 7 - b - 1 = 4

a - b = -2 ------(1)

Now we know line AB is perpendicular to the line x - y = 2

Therefore, by the property of two perpendicular lines,

$m_{1}\times m_{2}=(-1)$

x - y = 2

y = x - 2

Then slope of the line will be $m_{1}=1$

Now slope of line AB will be $m_{2}=-1$

Slope joining two points (7, 1) and (a, b) is (-1)

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

b - 1 = 7 - a

a + b = 8 --------(2)

By adding equations (1) and (2)

2a = 6

a = 3

and b = 8 - 3 = 5

Therefore, the given point (a, b) is (3, 5).

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