Question

Find the value of x for which the distance between the (4,-5) and (12,x) is 10 units

Answer 1

Answer:

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Answer 2

Answer:

x = -11, 1

Step-by-step explanation:

In the question,

We know that the distance between two points is given by,

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

We have the points,

(4, -5) and (12, x).

The distance between these two points is given as 10.

So,

Distance between these two points is given by,

$Distance=\sqrt{(x-(-5))^{2}+(12-4)^{2}}=10\\10=\sqrt{(x+5)^{2}+64}$

On squaring the both sides we get,

$10=\sqrt{(x+5)^{2}+64}\\100=x^{2}+25+10x+64\\x^{2}+10x-11=0\\(x+11)(x-1)=0\\x=-11,1$

Therefore, the possible values of 'x' for which the distance between the points is 10 units are,

x = -11, 1