Question

find the value of y if distance between two points A(2,-2) and B(-1,y) is 5.​

Answer 1

Answer:

Step-by-step explanation:

distance=[(x2-x1)^2+(y2-y1)]^1/2

here x1=2 x2=-1 y1=-2 y2=y distance=5

5=[(-1-1)^2+(y+2)^2]^1/2

5=[4+y^2+4+4y]^1/2

squaring on both side

25=4+y^2+4+4y

y^2+4y+8-25=0

y^2+4y-17=0

apply quadratic formula now

Answer 2

The value of y is -6 and 2.

Step-by-step explanation:

Given : If distance between two points A(2,-2) and B(-1,y) is 5.​

To find : The value of y ?

Solution :

The distance formula is $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Here, distance is d=5

The point is $(x_1,y_1)=A(2,-2)$ and $(x_2,y_2)=B(-1,y)$

Substitute the value,

$5=\sqrt{(-1-2)^2+(y-(-2))^2}$

$5=\sqrt{(-3)^2+(y+2)^2}$

Squaring both side,

$25=9+y^2+4+4y$

$y^2+4y-12=0$

$y^2+6y-2y-12=0$

$y(y+6)-2(y+6)=0$

$(y+6)(y-2)=0$

$y=-6,2$

Therefore, the value of y is -6 and 2.

#Learn more

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