Respuesta :
Given that line segment RU with vertices R(1, 1) and U(4, 5) is considered the base of parallelogram RSTU.
Then, the line segment ST with vertices S(7, 0) and T(10, 4) is the top of the parallelogram.
The corresponding height of the parallelogram is the length of a line with endponts at RU and ST and perpendicular to both RU and ST.
The equation of the line segment RU is given by
[tex] \frac{y-1}{x-1} = \frac{5-1}{4-1} = \frac{4}{3} \\ \\ 3(y-1)=4(x-1) \\ \\ 3y-3=4x-4 \\ \\ 3y=4x-1 \\ \\ y= \frac{4}{3} x- \frac{1}{3} [/tex]
Recall that given that two lines are perpendicular, the product of the slope of the two lines is -1.
Let the slope of the line perpendicular to line RU be m, then
[tex] \frac{4}{3} m=-1 \\ \\ m=- \frac{3}{4} [/tex]
Thus, the equation of the line perpendicular to RU passing through point (1, 1) is given by
[tex]y-1=- \frac{3}{4} (x-1) \\ \\ 4(y-1)=-3(x-1) \\ \\ 4y-4=-3x+3 \\ \\ 4y=-3x+7 \\ \\ y=- \frac{3}{4} x+ \frac{7}{4} [/tex]
The equation of the line segment ST is given by
[tex] \frac{y-0}{x-7} = \frac{4-0}{10-7} = \frac{4}{3} \\ \\ 3y=4(x-7)=4x-28 \\ \\ y= \frac{4}{3} x- \frac{28}{3} [/tex]
The line perpendicular to line segment RU intersected line segment ST at the point given by
[tex]- \frac{3}{4} x+ \frac{7}{4}=\frac{4}{3} x- \frac{28}{3} \\ \\ \frac{4}{3} x+\frac{3}{4} x=\frac{7}{4}+\frac{28}{3} \\ \\ \frac{25}{12} x= \frac{133}{12} \\ \\ x= \frac{133}{25} \\ \\ y=\frac{4}{3} \left(\frac{133}{25}\right)- \frac{28}{3}= -\frac{56}{25} [/tex]
Thus, the corresponding height of the parallelogram is the line with endpoints
[tex](1,1) \ and \ \left(\frac{133}{25},-\frac{56}{25}\right)[/tex]
Recall that the length of a line passing through points
[tex](x_1,y_1) \ and \ (x_2,y_2)[/tex]
is given by
[tex]l= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} [/tex]
Thus, the length of the line passing through points
[tex](1,1) \ and \ \left(\frac{133}{25},-\frac{56}{25}\right)[/tex]
is given by
[tex]l= \sqrt{\left(\frac{133}{25}-1\right)^2+\left(-\frac{56}{25}-1\right)^2} \\ \\ = \sqrt{\left( \frac{108}{25}\right)^2+\left(- \frac{81}{25} \right)^2}= \sqrt{ \frac{11,664}{625} + \frac{6,561}{625} } \\ \\ = \sqrt{ \frac{729}{25} } = \frac{27}{5} =5.4[/tex]
Therefore, the corresponding height of the given parallelogram is 5.4 units
Then, the line segment ST with vertices S(7, 0) and T(10, 4) is the top of the parallelogram.
The corresponding height of the parallelogram is the length of a line with endponts at RU and ST and perpendicular to both RU and ST.
The equation of the line segment RU is given by
[tex] \frac{y-1}{x-1} = \frac{5-1}{4-1} = \frac{4}{3} \\ \\ 3(y-1)=4(x-1) \\ \\ 3y-3=4x-4 \\ \\ 3y=4x-1 \\ \\ y= \frac{4}{3} x- \frac{1}{3} [/tex]
Recall that given that two lines are perpendicular, the product of the slope of the two lines is -1.
Let the slope of the line perpendicular to line RU be m, then
[tex] \frac{4}{3} m=-1 \\ \\ m=- \frac{3}{4} [/tex]
Thus, the equation of the line perpendicular to RU passing through point (1, 1) is given by
[tex]y-1=- \frac{3}{4} (x-1) \\ \\ 4(y-1)=-3(x-1) \\ \\ 4y-4=-3x+3 \\ \\ 4y=-3x+7 \\ \\ y=- \frac{3}{4} x+ \frac{7}{4} [/tex]
The equation of the line segment ST is given by
[tex] \frac{y-0}{x-7} = \frac{4-0}{10-7} = \frac{4}{3} \\ \\ 3y=4(x-7)=4x-28 \\ \\ y= \frac{4}{3} x- \frac{28}{3} [/tex]
The line perpendicular to line segment RU intersected line segment ST at the point given by
[tex]- \frac{3}{4} x+ \frac{7}{4}=\frac{4}{3} x- \frac{28}{3} \\ \\ \frac{4}{3} x+\frac{3}{4} x=\frac{7}{4}+\frac{28}{3} \\ \\ \frac{25}{12} x= \frac{133}{12} \\ \\ x= \frac{133}{25} \\ \\ y=\frac{4}{3} \left(\frac{133}{25}\right)- \frac{28}{3}= -\frac{56}{25} [/tex]
Thus, the corresponding height of the parallelogram is the line with endpoints
[tex](1,1) \ and \ \left(\frac{133}{25},-\frac{56}{25}\right)[/tex]
Recall that the length of a line passing through points
[tex](x_1,y_1) \ and \ (x_2,y_2)[/tex]
is given by
[tex]l= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} [/tex]
Thus, the length of the line passing through points
[tex](1,1) \ and \ \left(\frac{133}{25},-\frac{56}{25}\right)[/tex]
is given by
[tex]l= \sqrt{\left(\frac{133}{25}-1\right)^2+\left(-\frac{56}{25}-1\right)^2} \\ \\ = \sqrt{\left( \frac{108}{25}\right)^2+\left(- \frac{81}{25} \right)^2}= \sqrt{ \frac{11,664}{625} + \frac{6,561}{625} } \\ \\ = \sqrt{ \frac{729}{25} } = \frac{27}{5} =5.4[/tex]
Therefore, the corresponding height of the given parallelogram is 5.4 units
Height of the parallelogram is the shortest distance from the base of the parallelogram to the opposite vertex.
The corresponding height of the parallelogram is 5.4 units. Thus the option B is the correct option.
Given information-
Vertices for the given parallelogram are R(1,1), U(10,4) S(10,4) and T(7,0).
Line segment RU is base of parallelogram RSTU.
What is height of parallelogram?
Height of the parallelogram is the shortest distance from the base of the parallelogram to the opposite vertex.
Height of the given parallelogram is the perpendicular distance from the point S to the line segment RU.
Length of RU,
[tex]\begin{aligned}\\\dfrac{y-y_1}{y_2-y_1} &=\dfrac{x-x_1}{x_2-x_1} \\\dfrac{y-5}{1-5} &=\dfrac{x-4}{1-4} \\\dfrac{y-5}{-4} &=\dfrac{x-4}{-3}\\3(y-5)&=4(x-4)\\4x-3y-1&=0\\\end[/tex]
Drop a perpendicular SP from point S to line segment RU. Thus the pints of this perpendicular point P on above line segment can be find as,
[tex]d=\dfrac{|ax+by+c|}{\sqrt{a^2+b^2} } \\d=\dfrac{|4\times 7+(-3)\times0+(-1)|}{\sqrt{4^2+(-3)^2} } \\\\d=\dfrac{27}{\sqrt{25} } \\d=\frac{27}{5} \\d=5.4[/tex]
Hence the corresponding height of the parallelogram is 5.4 units. Thus the option B is the correct option.
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