Question

1) Find the value of y if distance between points AC 2,-2) & B (-1, y) is 5.​

Answer 1

Answer:

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

(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)(x

1

,y

1

)=A(2,−2) and (x_2,y_2)=B(-1,y)(x

2

,y

2

)=B(−1,y)

Substitute the value,

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

(−1−2)

2

+(y−(−2))

2

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

(−3)

2

+(y+2)

2

Squaring both side,

25=9+y^2+4+4y25=9+y

2

+4+4y

y^2+4y-12=0y

2

+4y−12=0

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

2

+6y−2y−12=0

y(y+6)-2(y+6)=0y(y+6)−2(y+6)=0

(y+6)(y-2)=0(y+6)(y−2)=0

y=-6,2y=−6,2

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