If the distance between the points
(3, a) and (6,1) is s, find the value of a
Answers
Answer:
The value of a is 5.
Step-by-step explanation:
Given : If the distance between the points (3,a) and (6,1) is 5.
To find : The value of a?
Solution :
The distance formula of two points 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, (x_1,y_1)=(3,a)(x
1
,y
1
)=(3,a) , (x_2,y_2)=(6,1)(x
2
,y
2
)=(6,1) and d=5
5=\sqrt{(6-3)^2+(1-a)^2}5=
(6−3)
2
+(1−a)
2
Squaring both side,
25=(3)^2+(1-a)^225=(3)
2
+(1−a)
2
25=9+1+a^2-2a25=9+1+a
2
−2a
a^2-2a-15=0a
2
−2a−15=0
a^2-5a+3a-15=0a
2
−5a+3a−15=0
a(a-5)+3(a-5)=0a(a−5)+3(a−5)=0
(a-5)(a+3)=0(a−5)(a+3)=0
a=5,-3a=5,−3
Reject a=-3.
The value of a is 5.
Answer:
The value of a is 5.
Step-by-step explanation:
Given : If the distance between the points (3,a) and (6,1) is 5.
To find : The value of a?
Solution :
The distance formula of two points 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, (x_1,y_1)=(3,a)(x
1
,y
1
)=(3,a) , (x_2,y_2)=(6,1)(x
2
,y
2
)=(6,1) and d=5
5=\sqrt{(6-3)^2+(1-a)^2}5=
(6−3)
2
+(1−a)
2
Squaring both side,
25=(3)^2+(1-a)^225=(3)
2
+(1−a)
2
25=9+1+a^2-2a25=9+1+a
2
−2a
a^2-2a-15=0a
2
−2a−15=0
a^2-5a+3a-15=0a
2
−5a+3a−15=0
a(a-5)+3(a-5)=0a(a−5)+3(a−5)=0
(a-5)(a+3)=0(a−5)(a+3)=0
a=5,-3a=5,−3
Reject a=-3.
The value of a is 5.