Given the line (1-n)*x + n*y -1 = 0 calculate n for the line is perpendicular to another line that passes through (-2,3) , (1,2).

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

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Let d1 and d2 be two perpendicular lines such that:

 d1 : (1-n)*x + n*y -1 = 0

d1: ny = -(1-n)x + 1

d1: y = [-(1-n)/n]x  + 1/n

==> the slope m1 = (n-1)/n

d2 passes through the points (-2, 3) and (1,2)

The, the slope m2= ((2-3)/(1--2) = -1/3

But we know thet d1 and d2 are perpendicular, then:

m1 * m2 = -1

(n-1)/n * -1/3 = -1

==> (n-1)/n = 3

==> n-1 = 3n

==> -1 = 2n

==> n= -1/2

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

We  shall determine the line through (-2,3) and (1,2).

The line through (x1,y1) and (x2,y2) is given by:

y -y1 = {(y2-y1)/(x2-x1)} (x-x1)...(1)

(x1 ,y1) = (-2,3) and (x2, y2) = 1,2). Substituting in the formula (1), we get:

(y-3) = {(2-3)/(1- -2)}(x --2)

y -3 = (-1/3) (x+2)

(3y-9) = (-(x+2)

 x+3y -9+2 = 0

x+3y -7 = 0.This is the line through the given 2 points.

Any line perpendicular to this is  got by reversing the coefficients of x and y and putting a minus sign to one of the coefficints.

So x+3y -7 = 0 has the perpendicular is of the form:

3x-y +k = 0. Comparing this line with the line (1-n)+ ny -1 = 0. If both equations  are of the same line then the coefficits of x , y and contant terms should have the same proportion:

(1-n)/3 = n/-1 = k/-1.

From the first two equations, (1-n)/3 = n/(-1)

n-1 = 3n

-1 = 3n-n

-1= 2n

n = -1/2.

So n = -1/2.

 

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

If the 2  lines are perpendicular, then the product of the values of their slopes is -1.

Since the given equation isn't put in the standard form, y = mx + n, we'll find it's slope:

(1-n)*x + n*y -1 = 0

n*y = -(1-n)*x + 1

We'll divide by n:

y = -(1-n)*x/n + 1/n, so the slope is m1 = (n-1)/n

That means that the second slope is:

m1*m2 = -1

m2 = -1/m1

m2 = -n/(n-1)

Now, we'll write the equation of a line that passe through the given points:

 (x2-x1)/(x-x1) = (y2-y1)/(y-y1)

We'll substitute the coordinates of the given points into the formula:

(1+2)/(x+2) = (2-3)/(y-3)

3/(x+2) = -1/(y-3)

We'll cross multiply:

-(x+2) = 3(y-3)

We'll remove the brackets:

-x-2 = 3y - 9

We'll put the equation into the standard form:

3y = -x-2+9

We'll divide by 3:

y = -x/3 + 7/3, so m2 = -1/3

But m2 = -n/(n-1) => -1/3 = -n/(n-1)

We'll cross multiply:

-3n = 1-n

-2n = 1

We'll divide by -2:

n  =-1/2

The equation of the line is: (1+ 1/2)*x -y/2 -1 = 0

or

3x - y - 2 = 0

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