SLOPES OF PARALLEL AND PERPENDICULAR LINES WITH UNKNOWN COORDINATES

Problem 1 :

A, B and C have coordinates (2, 9), (10, −7) and (6, k) respectively. AB is perpendicular to AC.

Solution :

Let the points be A (2, 9), B (10, −7) and C (6, k)

Lines AB and AC are perpendicular.

Slope of AB :

(2, 9), (10, −7)

Slope m = (y₂ - y₁) / (x₂ - x₁)

= (-7 - 9) / (10 - 2)

= -16/8

= -2 ------(1)

Slope of AC :

A (2, 9) and C(6, k)

Slope (m) = (k - 9) / (6 - 2) 

= (k - 9) / 4 ----------(2)

(1) x (2) = -1

(k - 9)/4 ⋅ (-2) = -1

(k - 9) / 2 = 1

k - 9 = 2

k = 2 + 9

k = 11

So, the value of k is 11.

Problem 2 :

Line A passes through the points (3, 6) and (5, -2) Line B passes through the points (2, 5) and (8, k) Line A and Line B are parallel. Find the value of k.

Solution :

Slope of the line A :

Slope of the line passes through the points (3, 6) and (5, -2)

= (-3 - 6) / (5 - 3)

= -9/2 -------(1)

Slope of the line B :

Slope of the line passes through the points (2, 5) and (8, k)

= (k - 5) / (8 - 2)

= (k - 5) / 6 ------(2)

Since the lines A and B are parallel, their slopes will be equal.

-9/2 = (k - 5) / 6

-9(6) = 2(k - 5)

-54 = 2k - 10

-54 + 10 = 2k

2k = -44

k = -44/2

k = -22

Problem 3 :

Line A passes through the points (-3, -1) and (-1, 9) Line B passes through the points (-2, 1) and (k, 4) Line A and Line B are perpendicular. Find the value of k.

Solution :

Slope of the line A :

Slope of the line passes through the points (-3, -1) and (-1, 9)

= (9 + 1) / (-1 + 3)

= 10/2

= 5-------(1)

Slope of the line B :

Slope of the line passes through the points (-2, 1) and (k, 4)

= (4 - 1) / (k + 2)

= 3/(k + 2) ------(2)

Since the lines A and B are perpendicular, the product of their slopes will be equal to -1.

5 ⋅ [3/(k + 2)] = -1

15/(k + 2) = -1

15 = -(k + 2)

-15 = k + 2

k = -15 - 2

k = -17

When the lines are perpendicular, the value of k is -17.

Problem 4 :

The line through (−1, k) and (−7, −2) is parallel to the line y = x + 1. Find a value for k based on the given description.

Solution :

Slope of the line passes through the points (-1, k) and (-7, -2)

= (-2 - k) / (-7 + 1)

= (-2 - k) / (-6)

Slope of the line y = x + 1

= 1

Since the lines are parallel, their slopes will be equal.

(-2 - k) / (-6) = 1

-2 - k = -6

-k = -6 + 2

-k = -4

k = 4

So, the value of k is 4.

Problem 5 :

The line through (k, 2) and (7, 0) is perpendicular to the line y = x − (28/5). Find a value for k based on the given description.

Solution :

Slope of the line passes through the points (k, 2) and (7, 0)

= (0 - 2) / (7 - k)

= (-2) / (7 - k)

Slope of the line y = x - (28/5)

= 1

Since the lines are perpendicular, the product of their slopes will be equal to -1.

(-2) / (7 - k) ⋅ 1  = -1

-2 = -(7 - k)

-2 = -7 + k

k = -2 + 7

k = 5

So, the value of k is 5.

Problem 6 :

If A (1, 3), B (–1, 2), C (2, 5) and D (x, 4) are the vertices of parallelogram ABCD then the value of x is

(a) 3        (b) 4     (c) 0            (d) 3/2

Solution :

In the shape parallelogram, the opposite sides will be parallel.

Slope of AB = (2 - 3) / ( -1 - 1)

= -1/(-2)

= 1/2

Slope of CD = (4 - 5) / (x - 2)

= -1/(x - 2)

1/2 = -1/(x - 2)

x - 2 = -2

x = -2 + 2

x = 0

So, the value of x is 0. Option c is correct.

Problem 7 :

A straight line is perpendicular to the straight line passing through (2, 8) and (6, 15) and passes through (0, 9) and (x, 17). Find the value of x.

Solution :

Slope of the line passes through the points (2, 8) and (6, 15)

= (15 - 8) / (6 - 2)

= 7/4

Slope of the line passes through the points (0, 9) and (x, 17)

= (17 - 9) / (x - 0)

= 8/x

Product of the slopes = -1

(7/4)(8/x) = -1

56/4x = -1

-4x = 56

x = 56/-4

x = -14

So, the value of x is -14.