FIND THE MISSING VALUE WITH PERPENDICULAR LINES

Example 1 :

Find the values of p for which the following two straight lines are perpendicular to each other.

8px + (2 - 3p)y + 1  =  0

px + 8y + 7  =  0

Solution :

If two lines are perpendicular then product of their slopes will be equal to -1.

Slope (m)  =  - coefficient of x/coefficient of y

Slope of the first line 8px + (2 - 3p)y + 1 = 0 is 

m₁  =  -8p / (2 - 3p)

Slope of the second line px + 8y + 7 = 0 is

m₂  =  -p/8

Because the lines are perpendicular, 

m₁ ⋅ m₂  =  -1

Substitute. 

[-8p / (2 - 3p)] ⋅ [-p/8]  =  -1

[(8p²) / 8(2 - 3p)]  =  -1

p² / (2 - 3p)  =  -1

p²  =  -1(2 - 3p)

p²  =  -2 + 3 p

p² - 3p + 2  =  0

(p - 2) (p - 1)  =  0

p - 2  =  0     (or)     p - 1  =  0

  p  =  2   (or)   p  =  1

Example 2 :

If the straight lines passing through the points (h, 3) and (4, 1) intersects the line 7x – 9y – 19 = 0 at right angle, then find the value of h.

Solution :

The required straight lines passing through the points (h, 3) and (4, 1) intersects the line 7x – 9y – 19 = 0 at right angle

Because the line joining the points (h, 3) and (4, 1) and the line 7x - 9y - 19  =  0 are perpendicular the product of their slopes will be equal to -1

Let m₁ be the slope of the line joining (h, 3) and (4, 1). 

Formula for slope of line joining two points is given by 

m  =  (y₂ - y₁) / (x₂ - x₁)

Then, 

m₁  =  (1 - 3) / (4 - h)

m₁  =  -2 / (4 - h)

Let m₂ be the slope of the line 7x – 9y – 19 = 0.

m₂  =  -7/(-9)

m₂  =  7/9

Because the lines intersect at right angle, they are perpendicular. 

Then, 

m₁ ⋅ m₂  =  -1

[-2 / (4 - h)] x (7 / 9)  =  -1

- 14 / 9(4 - h)  =  -1

Multiply each side by (-1).

- 14 / 9(4 - h)  =  -1

14 / (36 - 9h)  =  1

14  =  36 - 9h

Add 9h to each side. 

9h + 14  =  36

Subtract 14 from each side. 

h  =  22/9

Example 3 :

Given that P = (−1, −1), Q = (4, 3), A = (1, 2), and B = (7, k), find the value of k that makes the line AB

(a) parallel to PQ

(b) perpendicular to PQ

Solution :

a)  When the line AB is parallel to PQ, their slopes will be equal.

P = (−1, −1), Q = (4, 3)

Slope of PQ = (3 - (-1)) / (4 - (-1))

= (3 + 1)/(4 + 1)

= 4/5 ------(1)

Slope of AB = (k - 2) / (7 - 1)

= (k - 2) / 6 ------(2)

(1) = (2)

4/5 =  (k - 2)/6

6(4) = 5(k - 2)

24 = 5k - 10

5k = 24 + 10

5k = 34

k = 34/5

b) When the line AB is perpendicular to line PQ, the product of their slopes will be equal to -1.

(4/5) ⋅ (k - 2) / 6 = -1

2(k - 2) / 15 = -1

2k - 4 = -15

2k = -15 + 4

2k = -11

k = -11/2

Example 4 :

Let A = (−6, −4), B = (1, −1), C = (0, −4), and D = (−7, −7). Show that the opposite sides of quadrilateral ABCD are parallel. Such a quadrilateral is called a parallelogram.

Solution :

Slope of side AB = (-1 + 4) / (1 + 6)

= 3/7

Slope of BC = (-4 + 1) / (0 - 1)

= -3/(-1)

= 3

Slope of CD = (-7 + 4) / (-7 - 0)

= -3/(-7)

= 3/7

Slope of DA = (-7 + 4) / (-7 + 6)

= -3/(-1)

= 3

Slope of AB = Slope of CD

Slope of BC = Slope of DA

Since the opposite sides are having same slope, they are parallel and it is the shape of parallelogram.

Example 5 :

If the straight lines 12y = -(p + 3)x + 12, 12x -7y = 16 are perpendicular then find p.

Solution :

12y = -(p + 3)x + 12

y = -(p + 3)/12 + (12/12)

y = -(p + 3)x/12 + 1

m₁ = -(p + 3)/12

12x -7y = 16

7y = 12x - 16

y = (12/7)x - (16/7)

m₂ = 12/7

Because the lines are perpendicular, 

m₁ ⋅ m₂  =  -1

 -(p + 3)/12 ⋅ 12/7 = -1

(p + 3)/7 = 1

p + 3 = 7

p = 7 - 3

p = 4

So, the value of p is 4.

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