Question

इस द डिस्टेंस बिटवीन द पॉइंट (4,p) and (1,0) is 5 then p =​

Answer 1

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♧ Correct Que:-

=》 If the distance between the points (4,p) and (1,0) is 5 units then find the value of p.

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♧ ANSWER:-

▪︎ Given:-

• Two points (4,p) and (1,0)

• The distance between the point = 5 units.

▪︎ To Find:-

• The value of p.

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▪︎ Formula Used:-

• Distance Formula:-

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

Where,

• $\sf x_1 \:\:And \:\:x_2$ are x coordinates of given two points.

• $\sf y_1 \:\: And \:\:y_2$ are y coordinates of the given two points.

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▪︎ So Here,

• $\sf x_1\:\:And\:\:x_2$ are 4 and 1 respectively.

• $\sf y_1 \:\:And\:\:y_2$ are p and 0 respectively.

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▪︎ Therefore ATQ:-

$\tt\mapsto \: distance = \sqrt{(1 - 4) {}^{2} + (0 - p) {}^{2} } = 5. \\ \\ \tt\mapsto \sqrt{( - 3) {}^{2} + ( - p) {}^{2} } = 5 \\ \\ \tt \mapsto \sqrt{9 +p {}^{2} } = 5 \\ \\ \tt\mapsto( \sqrt{9 + p {}^{2} } ) {}^{2} = {5}^{2} \\ \\ \sf(squaring \: both \: sides) \\ \\ \tt\mapsto9 + p {}^{2} = 25 \\ \\ \tt\mapsto \: p^2 = 25 - 9 \\ \\ \tt\mapsto \: p {}^{2} = 16 \\ \\ \tt\mapsto \: p = \sqrt{16} \\ \\ \tt\mapsto \: p = 4.$

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$\therefore$The required value of p is 4.

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