Respuesta :
Answer:
a. perpendicular
b. perpendicular
c. perpendicular
Step-by-step explanation:
The condition for two lines to be perpendicular is:
[tex]m_{1}m_{2}=-1[/tex]
where [tex]m_{1}[/tex] is the slope of one line and [tex]m_{2}[/tex] the slope of the other line.
How we calculate the slope? if we have two points [tex](x_{1},y_{1})[/tex] and [tex](x_{2},y_{2})[/tex] the slope of the line between them is:
[tex]m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]
according to the problem the end point of all the lines for part a and par b is the origin, so [tex](x_{1},y_{1})[/tex] will be [tex](0,0)[/tex]
- part a,
lets calculate the slope of the line that passes between (0, 0) and (4,10), in this case
[tex]x_{1}=0\\y_{1}=0\\x_{2}=4\\y_{2}=10[/tex]
so the first slope is:
[tex]m_{1}=\frac{10-0}{4-0}=\frac{10}{4}[/tex]
and for the line that passes between the ponits (0,0 ) and (5, -2)
[tex]x_{1}=0\\y_{1}=0\\x_{2}=5\\y_{2}=-2[/tex]
so the second slope is:
[tex]m_{2}=\frac{-2-0}{5-0}=\frac{-2}{5}[/tex]
Now let's check if they are perpendicular:
[tex]m_{1}m_{2}=-1[/tex]
[tex](\frac{10}{4} )(\frac{-2}{5} )=\frac{-20}{20} =-1[/tex]
meets the condition, the lines in part a are perpendicular
- part b
we have the point: (-7,0) this is on the x-axis, the line with the origin is horizontal
and we have the point (0,-4) this is on the y-axis, the line with the origin is vertical.
Since one line is horizontal and the other vertical, the lines in part b are also perpendicular
- part c
the line between (-3,-2) and (1,8) has a slope:
[tex]m_{1}=\frac{8+2}{1+3}=\frac{10}{4}[/tex]
the line between (-3,-2) and the origin (2,-4) has a slope:
[tex]m_{2}=\frac{-4+2}{2+3}=\frac{-2}{5}[/tex]
Now let's check if they are perpendicular:
[tex]m_{1}m_{2}=-1[/tex]
[tex](\frac{10}{4} )(\frac{-2}{5} )=\frac{-20}{20}=-1 [/tex]
the lines are perpendicular.
