Respuesta :

a)


[tex]\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) Q&({{ 0}}\quad ,&{{ 2}})\quad % (c,d) P&({{ 0.5}}\quad ,&{{ 0}}) \end{array}\qquad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}[/tex]

[tex]\bf QP=\sqrt{(0.5-0)^2+(0-2)^2}\implies QP=\sqrt{0.5^2+2^2} \\\\\\ QP=\sqrt{\left( \frac{1}{2} \right)^2+4}\implies QP=\sqrt{ \frac{1^2}{2^2}+4}\implies QP=\sqrt{\frac{1}{4}+4} \\\\\\ QP=\sqrt{\frac{17}{4}}\implies QP=\cfrac{\sqrt{17}}{\sqrt{4}}\implies QP=\cfrac{\sqrt{17}}{2}[/tex]

b)

since QR=QP, that means that QO is an angle bisector, and thus the segments it makes at the bottom of RO and OP, are also equal, thus RO=OP

thus, since the point P is 0.5 units away from the 0, point R is also 0.5 units away from 0 as well, however, is on the negative side, thus R (-0.5, 0)


c)

what's the equation of a line that passes through the points (-0.5, 0) and (0,2)?

[tex]\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) Q&({{ 0}}\quad ,&{{ 2}})\quad % (c,d) R&({{ -0.5}}\quad ,&{{ 0}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{0-2}{-0.5-0}\implies \cfrac{-2}{-0.5}[/tex]

[tex]\bf m=\cfrac{\frac{-2}{1}}{-\frac{1}{2}}\implies \cfrac{-2}{1}\cdot \cfrac{2}{-1}\implies 4 \\\\\\ % point-slope intercept y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-2=4(x-0)\implies y=4x+2\\ \left. \qquad \right. \uparrow\\ \textit{point-slope form}[/tex]