Which is the equation of the line that is perpendicular to 4x-5y-1=0 and it passes through the point (2,1) ?

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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.

m1*m2=-1

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.

(4/5)*m2=-1

m2=-5/4

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:

(y-y1)=m(x-x1)

(y-1)=(-5/4)*(x-2)

We'll remove the brackets:

y - 1 + 5x/4 - 5/2 = 0

**y + 5x/4 - 7/2 = 0**

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**

as

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).

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