The distance between A 2-4 and B 3-8 is -A- -15 unitsB- -17 unitsC- 4 unitsD- 4-5 units

The correct option is B √(17) units ∵ Distance between points (x_1,y_1) and(x_2,y_2) is √((x_2−x_1)^2+(y_2−y_1)^2) AB^2 = (3−2)^2 + (8−4)^2 AB^2 = 1^2 + 4^2 A ...

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Standard X

Mathematics

Graphical Representation of Quadratic Equation

The distance ...

Question

The distance between A (2, 4) and B (3, 8) is :

4 units

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$\sqrt{17}$ units

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$\sqrt{15}$ units

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4.5 units

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Solution

The correct option is B
$\sqrt{17}$ units

$∵$ Distance between points $(x_{1}, y_{1}) a n d (x_{2}, y_{2}) i s \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

$A B^{2} = (3 - 2)^{2} + (8 - 4)^{2}$

$A B^{2} = 1^{2} + 4^{2}$

$A B^{2} = 1 + 16$

$A B^{2} = 17$

$A B = \sqrt{17}$ units.

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The distance between A (2, 4) and B (3, 8) is :

The correct option is B √17 units ∵ Distance between points (x1,y1)and(x2,y2) is √(x2−x1)2+(y2−y1)2 AB2=(3−2)2+(8−4)2 AB2=12+42 AB2=1+16 AB2=17 AB=√17units.

The correct option is B
$\sqrt{17}$ units

$∵$ Distance between points $(x_{1}, y_{1}) a n d (x_{2}, y_{2}) i s \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

$A B^{2} = (3 - 2)^{2} + (8 - 4)^{2}$

$A B^{2} = 1^{2} + 4^{2}$

$A B^{2} = 1 + 16$

$A B^{2} = 17$

$A B = \sqrt{17}$ units.