Question

The distance between point P(4,3)Q(a,–2)is5 find the value of a

Answer 1

Given:

• Points are : p (4,3) and Q(a,-2).

• Distance = 5 .

To find :

• The value of a.

Solution :

As we know that :

$\star \boxed{ \red{\bold{ d = \sqrt{ ({x_2 - x_1)}^{2} + {(y_2 - y_1)}^{2} } }}}$

• Here,

• x1= 4 and y1= 3.

• x2= a and y2= -2.

$\implies \: {(5)}^{2} = {(a - 4)}^{2} + {( - 2 - 3)}^{2} \\ \\ \implies \: 25 = {a}^{2} + 16 - 8a + 25\\ \\ \implies \: \: {a}^{2} - 8a \: + 16= 0 \\ \\ \implies \: {a}^{2} - 4a - 4a + 16 = 0 \\ \\ \implies \: a(a - 4) - 4(a - 4) = 0 \\ \\ \implies \: (a - 4)(a - 4) = 0 \\ \\ \implies \: a = 4 \:and \: 4$

Hence ,the value of a is 4.

Answer 2

Step-by-step explanation:

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