Question

2.
If the distance between the points (3,a) and (6,1) is 5. find the value of A.​

Answer 1

Answer:

Step-by-step explanation:

a=5 or -3,

√-3²+(a-1)²,

=5,

Squaring on both sides:

9+(a-1)^2=25,

(a-1)^2=16,

a-1=+4 or -4,

Hence a is 5 or -3.

Hope it helps.

Answer 2

Answer:

$\huge\underline{ \underline{ \red{your \: \: answer}}}$

Step-by-step explanation:

$\bf\pink{distannce \: formula \: d \mapsto \sqrt{ ({x2 - x1})^{2} + ( {y2 - y1})^{2} } } \\ \\ \bf\pink{let \: the \: points \: are \: p(3,a) ,Q(6,1)\: } \\ \\ \bf\pink{here \: pQ =5} \\ \\ \bf\pink{pQ= \sqrt{ ({x2 - x1})^{2} + ( {y2 - y1}^{2} ) } } \\ \\ \bf\pink{ \mapsto> 5 =\sqrt{{( 6 - 3})^{2} +( { 1 - a})^{2} } } \\ \\ \bf\pink{\mapsto > 5 = \sqrt{ 9 + 1 + {a}^{2} - 2a}} \\ \\ \bf\pink{ \mapsto> 25= { a}^{2} - 2a + 10} \\ \\ \bf\pink{ \mapsto > {a}^{2} - 2a - 35 =0 } \\ \\ \bf\pink{ \mapsto > {a}^{2} - (7a - 5a) - 35=0} \\ \\ \bf\pink{ \mapsto> {a}^{2} - 7a + 5a - 35 = 0 } \\ \\ \bf\pink{ \mapsto> a (a - 7) + 5(a - 7) =0} \\ \\ \bf\pink{ \mapsto > (a - 7)(a + 5) = 0} \\ \\ \bf\pink{a - 7 =0} \\ \bf\pink{ \mapsto > a =7} \\ \\ \bf\pink{a + 5 = 0} \\ \bf\pink{a = - 5} \\ \\ \bf\green{ points \: are \: (3,7)(6,1) \: or( 3, - 5)(6,1)} \\ \\ \large\boxed{ \fcolorbox{violet}{lime}{hope \:it \: help \: you} }$