Xy has endpoints at X(2,-3) and Y(-1,2) what is the length of XY

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Xy has endpoints at X(2,-3) and Y(-1,2) what is the length of XY

2 Units

Step-by-step explanation:

The length of xy = [tex]\sqrt{ (-1-2)^{2} + (2+3)^{2} } = \sqrt{ (-3)^{2} + 5^{2} } = \sqrt{9+25} = \sqrt{34} ;[/tex]

6

Step-by-step explanation:

The length form -1 to 2 is 3

The length from -3 to 2 is 5

√3²+5² = √9+25 =√36=6

Let x(2,-3) (x1 , y1) and y(-1 , 2) (x2 , y2). Using distance formula, lXYl = [tex]\sqrt{[2-(-1)] ^{2}+[-3-2] ^{2} } = \sqrt{1+25} = \sqrt{26} = 5.09[/tex]units.

Let x(2,-3) (x1 , y1) and y(-1 , 2) (x2 , y2). Using distance formula, lXYl = [tex]\sqrt{[2-(-1)] ^{2}+[-3-2] ^{2} } = \sqrt{1+25} = \sqrt{26} = 5.09[/tex]units.

[tex]\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) X&({{ 2}}\quad ,&{{ -3}})\quad % (c,d) Y&({{ -1}}\quad ,&{{2}}) \end{array}\qquad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2} \\\\\\ XY=\sqrt{[-1-2]^2+[2-(-3)]^2}\implies XY=\sqrt{(-1-2)^2+(2+3)^2} \\\\\\ XY=\sqrt{(-3)^2+5^2}\implies XY=\sqrt{34}[/tex]

The length of xy = [tex]\sqrt{ (-1-2)^{2} + (2+3)^{2} } = \sqrt{ (-3)^{2} + 5^{2} } = \sqrt{9+25} = \sqrt{34} ;[/tex]