The distance formula between two points $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ is given by

(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}

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(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

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

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None of the above

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Solution

The correct option is C $\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$
By the distance formula, the distance between the given two point si given by:
$A B = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$
So, option C is correct.

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are

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P and Q are 2 external points from which tangents are drawn to circle centered at origin and radius 'r' $P \equiv (x_{1}, y_{1}), Q \equiv (x_{2}, y_{2}) .$ What is the condition for the lengths of tangents to be the same?

(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2} = a^{2}

;

(x_{2} - x_{3})^{2} + (y_{2} - y_{3})^{2} = b^{2}

;

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;
then find

4 {∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & 1 x_{2} & y_{2} & 1 x_{3} & y_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣}^{2} =

Q. If

(x_{1}, y_{1}) = (2, 4)

and

(x_{2}, y_{2}) = (1, - 2)

, then find

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

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The distance formula between two points A(x1,y1) and B(x2,y2) is given by

The correct option is C √(x2−x1)2+(y2−y1)2By the distance formula, the distance between the given two point si given by:AB=√(x2−x1)2+(y2−y1)2So, option C is correct.

The distance formula between two points $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ is given by

The correct option is C $\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$
By the distance formula, the distance between the given two point si given by:
$A B = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$
So, option C is correct.