Hallar el area de un triangulo equilatero cuyas bases miden 70dm y 134 dm y el lado oblicuo 85dm
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Answer:
slope of parallel line and perpendicular line are 5 and -1/5 espectively
equation of parallel and perpendicular line are y = 5x + 22 respectively
Step-by-step explanation:
y = 5x - 4 is in the form
y = mx + c
where m is the slope of the line and c is the y intercept of thr line
therefore slope of the line = 5 and y intercept = -4
when an another line is parallel to the given line then the slope of both the lines are equal
therefore the slope the parallel line = 5
equation of a line passing through a given point with slope m is given by
given = (-4,2)
therefore equation of line y-2 = 5(x+4)
therefore y = 2+ 5x+20
y = 5x + 22is the eqaution of required line.
when two lines are perpendiculer then
where are slope of the lines therefore
m×5=-1
therefore m=
therefore eqaution of line passing through (-4,2) and with slope m= is given by