Find the equation that is perpendicular to the line y=3, passing through the point -4,-3. With explanation how to solve.?
回答 (6)
2017-02-17 9:35 am
Since y = 3 is a horizontal line, then the line perpendicular to the line y = 3 is a vertical line.
Equation for the vertical line : x = c
where c is a constant.
Since (-4, -3) lies on x = c, then substitute x = -4 into x = c :
c = -4
The equation of the required line :
x = -4
x + 4 = 0
Equation for the vertical line : x = c
where c is a constant.
Since (-4, -3) lies on x = c, then substitute x = -4 into x = c :
c = -4
The equation of the required line :
x = -4
x + 4 = 0
2017-02-17 9:31 am
x = -4
2017-02-17 9:33 am
y = 3 is a horizontal line
The perpendicular will be a vertical line.
Since it passes through (-4,-3), the equation of the perpendicular is x = -4
The perpendicular will be a vertical line.
Since it passes through (-4,-3), the equation of the perpendicular is x = -4
2017-02-17 9:31 am
Line L, say, y = 3 is // to the x-axis. Any line, M, say, perpendicular to L,
must be // to the y-axis and be of form x = constant. If M passes through
(-4,-3) the constant must = -4. Therefore M's eqn is x = -4.
must be // to the y-axis and be of form x = constant. If M passes through
(-4,-3) the constant must = -4. Therefore M's eqn is x = -4.
2017-02-17 9:23 am
x=-4
2017-02-17 9:28 am
For a line to be perpendicular, the 2 slopes multiplied together need to = -1. So since the slope in the given equation is 0, you can not have a perpendicular line. No matter what you multiply, it will = 0, not -1. I would say no solution.
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