Question

Point P(K, 3) is the mid point of the line segment AB. If AB √52 If units and coordinates of A are (–3, 5), then find the value of K.

Answer 1

GIVEN :–

• A point P(k , 3) is the midpoint of Line segment AB.

• Length AB = √52 unit.

• Coordinate of point A (-3 , 5).

TO FIND :–

• Value of k = ?

SOLUTION :–

• According to the given condition –

$\\ \implies{ \bold{AP = \dfrac{ AB}{2} }}$

$\\ \implies{ \bold{AP = \dfrac{ \sqrt{52} }{2} }}$

$\\ \implies{ \bold{AP = \dfrac{ 2 \sqrt{13} }{2} }}$

$\\ \implies{ \bold{AP = \sqrt{13} }}$

• Distance between Point A(-3 , 5) and P(k , 3) is √13 unit.

☞ Distance between two points P(a , b) and Q(c , d) is –

$\\ \longrightarrow \: \: \large{ \boxed{ \bold{PQ = \sqrt{ {(c - a)}^{2} + {(d - b)}^{2} } }}}$

• So that –

$\\ \implies \: \: { \bold{AP = \sqrt{ {(k - ( - 3))}^{2} + {(3 - 5)}^{2} } }}$

$\\ \implies \: \: { \bold{ \sqrt{13} = \sqrt{ {(k - ( - 3))}^{2} + {(3 - 5)}^{2} } }}$

• Now square on both side –

$\\ \implies \: \: { \bold{ {13} = { {(k + 3)}^{2} + {( - 2)}^{2} } }}$

$\\ \implies \: \: { \bold{ {13} = { {(k + 3)}^{2} + 4} }}$

$\\ \implies \: \: { \bold{ { {(k + 3)}^{2} = 13 - 4} }}$

$\\ \implies \: \: { \bold{ { {(k + 3)}^{2} = 9} }}$

$\\ \implies \: \: { \bold{ { k + 3 = \sqrt{9}} }}$

$\\ \implies \: \: { \bold{ { k + 3 = \pm \: 3 } }}$

$\\ \implies \: \: { \bold{ { k = \pm \: 3 - 3 } }}$

• Take positive(+) sign :–

$\\ \implies \: \: { \bold{ { k = \: 3 - 3 } }}$

$\\ \implies \: \large { \boxed{ \bold{ { k = \: 0 } }}}$

• Take negative(-) sign :–

$\\ \implies \: \: { \bold{ { k = \: - 3 - 3 } }}$

$\\ \implies \: \large { \boxed{ \bold{ { k = -6 } }}}$