Find the distance between the points P -1-1 and Q -4-3- A- -13- -10C- 4D- -15

The correct option is A √(13)According to the Distance Formula : d(P,Q) = (x2 x1)2+(y2 y1)2 Here, x2= 4, x1= 1, y2=3, y1=1 d(P,Q) = √(( 4 ( 1))^2 + (3 1 ...

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Standard X

Mathematics

Distance Formula

Find the dist...

Question

Find the distance between the points P(-1 , 1) and Q(-4 , 3).

\sqrt{13}

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\sqrt{10}

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\sqrt{15}

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Solution

The correct option is A $\sqrt{13}$
According to the Distance Formula : d(P,Q) = $\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$
Here, $x_{2} = - 4, x_{1} = - 1, y_{2} = 3, y_{1} = 1$
d(P,Q) = $\sqrt{(- 4 - (- 1))^{2} + (3 - 1)^{2}}$
= $\sqrt{(- 3)^{2} + (2)^{2}}$
= $\sqrt{9 + 4}$
= $\sqrt{13}$

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Find the distance between the points P(-1 , 1) and Q(-4 , 3).

The correct option is A √13According to the Distance Formula : d(P,Q) = √(x2 − x1)2+(y2 − y1)2 Here, x2=−4, x1=−1, y2=3, y1=1 d(P,Q) = √(−4−(−1))2+(3−1)2 = √(−3)2+(2)2 = √9+4 = √13

The correct option is A $\sqrt{13}$
According to the Distance Formula : d(P,Q) = $\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$
Here, $x_{2} = - 4, x_{1} = - 1, y_{2} = 3, y_{1} = 1$
d(P,Q) = $\sqrt{(- 4 - (- 1))^{2} + (3 - 1)^{2}}$
= $\sqrt{(- 3)^{2} + (2)^{2}}$
= $\sqrt{9 + 4}$
= $\sqrt{13}$