Question

a point on line x-y+1=0 at a distance 2√2 from the point (1,2) is​

Answer 1

Answer:

Step-by-step explanation:

this can be solved by algebraic parameter

let x=t

y=1+t

distance between them =

(t-1)^2+(t-1)^2=8

t-1=+-2

t=3 and -1

therefore points are 3,4 and -1,0

Answer 2

Answer:

The point is (3, 4) or (-1, 0)

Step-by-step explanation:

Given the line is x-y+1 =0

Distance is $2\sqrt{2}$ and one point is  (1, 2)

Let the point (h, k) is satisfies the equation h-k+1 = 0.

then h-k+1 = 0

k = h +1

Therefore the pointe are (h, k) and (1, 2)

Now we find the distance

The two points (x1,y1) (x2,y2) then the distance between these two points is

$d = \sqrt{(x2-x1)^{2} +(y2-y1)^{2} }$

In this question the distance is given

$d=2 \sqrt{2}$

Now distance between (h, k) and (1, 2) is $2\sqrt{2}$

$d =\sqrt{(h-1)^{2}+(k - 2)^{2} }$

$2\sqrt{2} =\sqrt{(h-1)^{2}+(k-2)^{2} }$

put k = h +1 in above expression

$2\sqrt{2} =\sqrt{(h-1)^{2}+(h+1-2)^{2} } \\2\sqrt{2} =\sqrt{(h-1)^{2}+(h-1)^{2} }$

Squaring on both sides

$(2\sqrt{2})^{2} = (\sqrt{(h-1)^{2}+(h-1)^{2} } )^{2}$

$8 = 2(h-1)^{2} \\4 = (h-1)^{2}$

Therefore, h-1 = ±2

h - 1 = 2

h = 3

then k = h + 1

        k = 3 + 1

∴k = 4

The point (h, k) is (3, 4)

and h - 1 = -2

h = -1

k = h + 1

k = -1+1 = 0

The point (h, k) is (-1, 0)