for the implicit integration I got:
the vertical tangent would at points at which y' goes to infinity.
that happen when the right-hand side has singularities, that is, the denominator is 0:
so the two points y above have a vertical tangent.
What is an equation of the line that passes through the point(-1,-3) and is perpendicularly to the line x-2y=14
2 answers:
Answer:
2x + y + 5 = 0
Step-by-step explanation:
The equation of line is x-2y = 14 and we need to find the equⁿ of line which passes through (-1,-3) and is perpendicular to given line.
- Find the slope of given line by converting it into slope intercept form.
=> x - 2y = 14
=> 2y = x - 14
=> y = x-14/2
=> y = x/2 - 7
- Comparing to y = mx + c ,
=> Slope (m) = 1/2 .
Now the slope of line perpendicular to it , will have a slope ,
=> = -1/m
=> = -1/½
=> = -2
- Using point slope form ,
=> m = (y - y₁) / ( x - x₁)
=> -2 = { y - (-3) } / { x - (-1) }
=> -2 = y + 3/x + 1
=> -2(x + 1) = y + 3
=> -2x -2 = y + 3
=> y + 2x +3+2 = 0
=> 2x + y + 5 = 0
<h3><u>Hence </u><u>the</u><u> </u><u>equation</u><u> of</u><u> </u><u>line </u><u>perpendicular</u><u> to</u><u> it</u><u> </u><u>is </u><u>2</u><u>x</u><u> </u><u>+</u><u> </u><u>y </u><u>+</u><u> </u><u>5</u><u> </u><u>=</u><u> </u><u>0</u><u> </u><u>.</u></h3>
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