Question

। (If the distance between points P(3.a) and Q(3, 1) is 4 units then find the value of a.)। (If the distance
between points P(3.a) and Q(3, 1) is 4 units then find
the value of a.)​

Answer 1

Answer:

Solution

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Let A(3,a) and B(4,1) be the points of a line AB

Distance AB=

(x

2

−x

1

)

2

+(y

2

−y

1

)

2

⇒

10

=

(1−a)

2

+(4−3)

2

Squaring both sides, we get

⇒10=(1−a)

2

+(4−3)

2

⇒10=1+a

2

−2a+1=0

⇒a

2

−2a−8=0

⇒a

2

−4a+2a−8=0

⇒a(a−4)+2(a−4)=0

⇒(a−4)(a+2)=0

∴a=−2 or a=4

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