Question

The distance between two points, P(-3, -2) and Q(b, 2b), is 10.
Find the two possible values of b.​

Answer 1

Required Answer:-

Given:

• Two points, P(-3, -2) and, Q(b, 2b)
• Distance between P and Q is 10 units.

To find:

• Two possible values of b.

Solution:

Distance between two points is given by the formula,

$\rm D = \sqrt{ {(x_{2} - x_{1}) }^{2} + {(y_{2} - y_{1})}^{2} }$

Here,

$\rm \mapsto x_{1} = - 3$

$\rm \mapsto x_{2} =b$

$\rm \mapsto y_{1} = - 2$

$\rm \mapsto y_{2} =2b$

So, according to the given condition,

$\rm \implies \sqrt{ {(b - ( - 3))}^{2} + (2b - ( - 2))^{2} } = 10$

$\rm \implies \sqrt{ {(b + 3)}^{2} + (2b + 2)^{2} } = 10$

Squaring both sides, we get,

$\rm \implies {(b + 3)}^{2} + (2b + 2)^{2} = 100$

$\rm \implies {b}^{2} + 6b + 9 + 4 {b}^{2} + 8b + 4= 100$

$\rm \implies 5{b}^{2} + 14b +13= 100$

$\rm \implies 5{b}^{2} + 14b +13 - 100 = 0$

$\rm \implies 5{b}^{2} + 14b - 87 = 0$

$\rm \implies 5{b}^{2} - 15b + 29b - 87 = 0$

$\rm \implies 5b(b - 3)+ 29(b - 3) = 0$

$\rm \implies(5b + 29)(b - 3) = 0$

By zero product rule,

Either 5b + 29 = 0 or b - 3 = 0

So,

$\rm \implies 5b + 29 = 0$

$\rm \implies b = \dfrac{ - 29}{5}$

$\rm \implies b = - 5.8$

Also,

$\rm \implies b - 3= 0$

$\rm \implies b = 3$

Hence, the possible values of b are -5.8 and 3.

Answer:

• The two possible values of b are -5.8 and 3.

Answer 2

Answer:

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