Question

if the distance between the points A(2,-3) and B (2,a) is 5, then the negative value of a is​

Answer 1

Question :-

If the distance between the points A(2,-3) and

B(2,a) is 5 units ,then find the negative value of a.

Answer :-

→ a = 2 .

or may be

$\to \boxed{\sf a = \frac{ - 6 \pm \sqrt{ - 28} }{2} } \\ \sf \: this \: is \: an \: imaginary \: number \: \:$

To Find :-

→ Negative value of a .

Solution :-

According to the question ,

→ Distance between A(2,-3) and B(2,a) is 5 units .

We know that ,

$\sf \: distance \: between \: points \: A(x_1 , y_1 ) \: \: and \: \\ \sf \: B(x_2 , y_2) \: is \: - \\ \\ \to \sf \: AB = \sqrt{ {(x_2 - x_1 )}^{2} + {(y_2 - y_1 )}^{2} }$

Given that AB = 5 units .

apply the distance formula between points A(2,-3)

and B(2,a)

$\to \sf AB = 5 = \sqrt{ {( 2 - 2 )}^{2} + { \{a - ( - 3) \}}^{2} } \\ \: \\ \to \sf \: 5 = \sqrt{ {(a + 3)}^{2} } \: \: ...eq.(1)\\ \\ \large \mathfrak{squaring \: on \: both \: sides \: } \\ \: \\ \to \sf \: 25 = {(a + 3)}^{2} \\ \\ \to \sf 25 = {a}^{2} + 9 + 6a \: \\ \\ \to \sf \: {a}^{2} + 6a + 9 - 25 = 0 \\ \\ \to \sf \: {a}^{2} + 6a + 16 = 0 \\ \\ \bf apply \: shri \: dharacharya \: principle \\ \\ \to \sf \: a = \frac{ - 6 \pm \sqrt{ {6}^{2} - 4 \times 16} }{2} \\ \\ \to \sf \: a = \frac{ - 6 \pm \sqrt{36 - 64} }{2} \\ \\ \to \boxed{\sf a = \frac{ - 6 \pm \sqrt{ - 28} }{2} } \\ \sf \: this \: is \: an \: imaginary \: number \: \\ \\ \sf \: or \: \\ \bf \: from \: eq.(1) \\ \\ \to \sf \: 5 = \sqrt{ {(a + 3)}^{2} } \\ \\ \to \sf 5 = a + 3 \\ \\ \to \boxed{\sf \: a = 2} \\ \\ \sf so \: here \: no \: value \: of \: a \: is \: negative \:$

Now ,check that 2 is the value of a or not .

Hence put a = 2

$\to \sf \: 5 = \sqrt{ {(2 + 3)}^{2} } \\ \\ \to \sf 5 = \sqrt{25} \\ \\ \to \: 5 = 5$

hence here a = 2 only .