Question

Find the value of y for which the distance between
the points P(2, 3) and O(10, y) is 10 units.​

Answer 1

AnsWer :

y = 9 and y = - 3

GiveN :

• The distance of point P and O is 10 units.

To FinD :

The value of y.

SolutioN :

We have,

• Point P ( 2 , 3 )
• Point O ( 10 , y )

✎ We know, Distance Formula.

$\tt \dagger \: \: \: \: \: D = \sqrt{ {\Big(x_2 - x_1 \Big)}^{2}+{ \Big(y_2-y_1\Big) }^{2} }$

$\tt : \implies10 = \sqrt{ {\Big(10 - 2 \Big)}^{2}+{ \Big(y-3\Big) }^{2} }$

$\tt : \implies10 = \sqrt{ {\Big(8 \Big)}^{2}+{ \Big(y-3\Big) }^{2} }$

$\tt : \implies10 = \sqrt{ 64+ {y}^{2} + 9 - 6y }$

$\tt : \implies10 = \sqrt{{y}^{2}- 6y + 73}$

$\tt : \implies10 0= {y}^{2}- 6y + 73$

$\tt : \implies0= {y}^{2}- 6y - 27$

$\rule{200}2$

☛ Let's Prime factor of ( 27 )

• Such that their product is - 27.
• And add / substrate is - 6.

$\begin{array}{r | l} 3 & 27 \\ \cline{2-2} 3 & 9 \\ \cline{2-2} 3 & 3 \\ \cline{2-2} & 1 \end{array}$

$\tt : \implies0= {y}^{2}- 9y + 3y - 27$

$\tt : \implies0= y(y - 9) + 3(y - 9)$

$\tt : \implies0= (y + 3)(y - 9)$

$\rule{90}1$

» Either,

$\tt : \implies y - 9 = 0.$

$\tt : \implies y = 9.$

$\rule{90}1$

» Or,

$\tt : \implies y + 3 = 0.$

$\tt : \implies y = - 3.$

✡ Therefore, the value of y is - 3 and 9.