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Which is the equation of the line that is perpendicular to 4x-5y-1=0 and it passes...
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We'll put the equation of the line into the standard form:
y=mx+n, where m is the slope of the line and n is the y intercept.
We'll put the equation in the standard form so we could use the property which says that the product between the slopes of 2 perpendicular lines is :-1
Let's suppose that the 2 slopes are m1 and m2.
We'll determine m1 from the given equation of the line, that is perpendicular to the one with the unknown equation.
The equation is 4x-5y-1=0.
We'll isolate -5y to the left side. For this reason, we'll subtract 4x-1 both sides:
-5y = -4x + 1
We'll multiply by -1:
5y = 4x - 1
We'll divide by 5:
y = 4x/5 - 1/5
The slope m1 = 4/5.
We also know that the line passes through the point (2,1), so the equation of a line which passes throuh a given point and it has a well known slope is:
We'll remove the brackets:
y - 1 + 5x/4 - 5/2 = 0
y + 5x/4 - 7/2 = 0
Posted by giorgiana1976 on September 30, 2010 at 12:32 AM (Answer #1)
We can write the equation 4x -5y -1 = 0 in the form
4x -5y -1 = 0
=> 5y = 4x -1
=> y = (4/5)x - 1/5
here the slope of the line is 4/5 . Now the product of the slopes or two perpendicular lines is -1.
So the slope of the line perpendicular to 4x -5y -1 = 0 is -5/4
Also the perpendicular passes through (2,1).
Hence we get the equation of the line as ( y -1) / (x -2)= -5/4
=> 4( y-1) = -5 (x-2)
=> 4y -4 = -5x + 10
=> 5x + 4y -4 -10 =0
=> 5x + 4y -14 = 0
The equation of the required line is 5x + 4y -14 = 0
Posted by william1941 on September 30, 2010 at 12:37 AM (Answer #2)
High School Teacher
The line perpendicular to the line ax+by+c = 0 is bx-ay +k.
So the line perpendicular to 4x-5y-1 = 0 is of the form 5x +4y+k = 0.
Since the line 5x+4y +k should pass through (2,1), the coordinates (2,1) should satisfy 5x+4y+k = 0.
Therefore, 5*(2)+4(1)+k = 0. So k = -10-4 = -14.
Therefore 5x+4y-14 = 0 is the line which is perpendicular to 4x-5y-1 = 0 and passes through the point (2,1).
Posted by neela on September 30, 2010 at 12:38 AM (Answer #3)
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