Question

if the distance between the point(0,y) and (8,9) is 8 unit ,find the value of y.

Answer 1

Answer:

Step-by-step explanation:

A(0,y)

B(8,9)

As,

distance=√(x'-x)^2+(y'-y)^2

x'=0 & x=8,y'=y & y=9

8=√(0-8)^2+(y-9)^2

(8)^2=64+y^2 + 81 - 18y

64=64+y^2+8-18y

0=y^2-18y+81

0=y^2-9y-9y+81

0=y(y-9)+9(y-9)

y+9=0

y-9=0

y=-9

y=9

Answer 2

let p (0,y) and Q (8,9)
and distance between pAnd Q =8
PQ=
$pq = \sqrt{(x2 - x1) { }^{2} + (y2 - y1) ^{2} }$

$8 = \sqrt({8- 0) {}^{2} } + (9 - y) ^{2}$
64=64+(81+y^2-18y)
y^2-18y +81=0
y^2-9y-9y+81=0
y (y-9)-9 (y-9)=0
therefore y=9