Question

(2) If the distance between the points (2, -3)
and (5. b) is 5, then b = ??
(step by step explaination needed)
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Answer 1

Answer:

b = 1, -7

Step-by-step explanation:

Here, AB = 5cm.

    $\setlength{\unitlength}{1cm}\begin{picture}(5, 5)\put(4.1, -2.3){\sf 5cm}\put(2, -2.5){\line(5,0){5}}\put(1.9, -2.6){ < }\put(2, -3){\line(5, 0){5}}\put(1.9, -3.06){\textbullet}\put(1.3, -3.5){\sf A(2, -3)}\put(6.9, -3.06){\textbullet}\put(6.8, -2.6){ > }\put(6.4, -3.5){\sf B(5, b)}\end{picture}$

We can use the Distance formula to find the unknown coordinate.

$\boxed{\sf Distance \ between \ any \ two \ points = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}}$

$\Longrightarrow \sf Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$\Longrightarrow \sf 5 = \sqrt{(2 - 5)^2 + (-3 - b)^2}$

$\Longrightarrow \sf 5 = \sqrt{(-3)^2 + (-3 - b)^2}$

$\Longrightarrow \sf 5 = \sqrt{9 + (-3 - b)^2}$

Square on both sides.

$\Longrightarrow \sf \Big(5\Big)^2 = \Big(\sqrt{9 + (-3 - b)^2} \Big)^2$

$\Longrightarrow \sf 25 = 9 + (-3 - b)^2$

Using (a - b)² = a² + b² - 2ab we get:

$\Longrightarrow \sf 25 = 9 + (-3)^2 + (b)^2 -2(-3)(b)^2$

$\Longrightarrow \sf 25 - 9 = 9 + b^2 + 6b$

$\Longrightarrow \sf 16 - 9 = b^2 + 6b$

$\Longrightarrow \sf b^2 + 6b - 7 = 0$

Splitting the middle term we get:

$\Longrightarrow \sf b^2 + 7b - b - 7 = 0$

$\Longrightarrow \sf b(b + 7) -1(b + 7) = 0$

$\Longrightarrow \sf (b - 1)(b + 7) = 0$

Equating the factors with zero we get:

$\Longrightarrow \sf b - 1 = 0$

$\Longrightarrow \boxed{ \sf b = 1}$

$\Longrightarrow \sf b + 7 = 0$

$\Longrightarrow \boxed{\sf b = - 7}$

Now let's check if both the values of b are valid ones.

Case I

If b = 1

$\Longrightarrow \sf Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$\Longrightarrow \sf 5 = \sqrt{(2 - 5)^2 + (-3 - 1)^2}$

$\Longrightarrow \sf 5 = \sqrt{(-3)^2 + (-4)^2}$

$\Longrightarrow \sf 5 = \sqrt{9 + 16}$

$\Longrightarrow \sf 5 = \sqrt{25}$

$\Longrightarrow \sf 5 = 5$

LHS = RHS

Hence b = 1.

Case II

If b = -7

$\Longrightarrow \sf Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$\Longrightarrow \sf 5 = \sqrt{(2 - 5)^2 + (-3 - (-7))^2}$

$\Longrightarrow \sf 5 = \sqrt{(-3)^2 + (-3 + 7)^2}$

$\Longrightarrow \sf 5 = \sqrt{(-3)^2 + (4)^2}$

$\Longrightarrow \sf 5 = \sqrt{9 + 16}$

$\Longrightarrow \sf 5 = \sqrt{25}$

$\Longrightarrow \sf 5 = 5$

LHS = RHS

Hence b = -7.

Therefore both the values, 1 and -7 are possible.

Answer 2

Given:-

❐Given Points = (2,-3) and (5,b)

❐Distance between given points = 5

Find:-

❐Value of b

Solution:-

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\linethickness{0.3mm}\qbezier(0,1)(1,1)(5.2,1)\put( - 0.2,0.7){\bf{A}} \put(5.2,0.7){\bf{B}} \put(5,0.4){\bf{(2,-3)}}\put( - 0.4,0.4){\bf{(5,b)}}\put(2.9, 1.3){\vector(1,0){2}} \put(2,1.3){\vector(-1,0){1.9}} \put(2.3,1.2){\bf{5}}\end{picture}$

we, know that

$\large{\underline{ \boxed{\sf Distance ,AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \: }}}$

where,

• AB = 5units
• $\sf x_1 = 2$
• $\sf x_2 = 5$
• $\sf y_1 = -3$
• $\sf y_2 = b$

So,

$\dashrightarrow\sf AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$\dashrightarrow\sf 5 = \sqrt{( 5 - 2)^2 + (-3 - b)^2}$

$\dashrightarrow\sf 5 = \sqrt{(3)^2 + (-3 - b)^2}$

$\dashrightarrow\sf 5 = \sqrt{9+ (-3- b)^2}$

$\sf \quad \bigstar using \: {(a - b)}^{2} = {a}^{2} - 2ab + {b}^{2} \bigstar$

$\dashrightarrow\sf 5 = \sqrt{9 + {-3}^{2} - 2(-3)(b) + {b}^{2} }$

$\dashrightarrow\sf 5 = \sqrt{9 + 9 + 6b+ {b}^{2} }$

$\dashrightarrow\sf 5 = \sqrt{18 +6b+ {b}^{2} }$

$\sf \qquad \bigstar both \: side \: squaring \bigstar$

$\dashrightarrow\sf 5^{2} = \bigg \{\sqrt{18 +6b+ {b}^{2} } \bigg \}^{2}$

$\dashrightarrow\sf 25 = 18 +6b+ {b}^{2}$

$\dashrightarrow\sf 18 - 25+6b+ {b}^{2} = 0$

$\dashrightarrow\sf -7+6b+ {b}^{2} = 0$

$\dashrightarrow\sf {b}^{2} + 6b - 7 = 0$

$\sf \quad \bigstar using \: middle - split \: term \bigstar$

$\dashrightarrow\sf {b}^{2} + 7b - b - 7 = 0$

$\dashrightarrow\sf b(b+ 7) - 1(b + 7 )= 0$

$\dashrightarrow\sf (b+ 7)(b - 1)= 0$

$\begin{gathered}\sf b+ 7 = 0 \\ \sf b = - 7 \end{gathered} \qquad\begin{gathered} \sf b - 1= 0 \\ \sf b = 1 \end{gathered}$

Hence, b = -7 or 1