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IF POINTS (A,-11) , Q (5,B ) , R(2,5) ,s(1,1) RE THE VERTICES OF THE PRALLELOGRAM PQRS THEN FIND THE VALUE OF A AND B

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Explanation:

let me write the information in clear text :

P = (A, -11)

Q = (5, B)

R = (2, 5)

S = (1, 1)

this is a parallelogram. that means every pair of opposite sides must be parallel. and that means they must have the same slope.

the slope of a line is the ratio

y coordinate difference / x coordinate difference

when going from one point on the line to another.

the slope of RS is

(5 - 1)/(2-1) = 4/1 = 4

the slope of PQ must be the same.

(-11 - B)/(A - 5) = 4

-11 - B = 4(A - 5) = 4A - 20

-B = 4A - 9

B = -4A + 9

and then 2 other 2 sides must be parallel (have the same slope) :

slope PR = slope QS

or

slope PS = slope QR

let's face it, there are 2 solutions for that reason, as we don't know which vertices are connected to which.

slope PR and QS :

(-11 - 5)/(A - 2) = (B - 1)/(5 - 1)

-16×4 = (B - 1)(A - 2) = AB - 2B - A + 2

-64 = AB - 2B - A + 2

0 = AB - 2B - A + 66

now, let's use our first identity from the slope of PQ in here :

0 = A(-4A + 9) - 2(-4A + 9) - A + 66

0 = -4A² + 9A + 8A - 18 - A + 66 = -4A² + 16A + 48

0 = -A² + 4A + 12 = A² - 4A - 12

remember the general solution of a quadratic equation

ax² + bx + c = 0

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

a = 1

b = -4

c = -12

x = A

A = (4 ± sqrt((-4)² - 4×1×-12))/(2×1) =

= (4 ± sqrt(16 + 48))/2 = (4 ± sqrt(64))/2 =

= (4 ± 8)/2 = 2 ± 4

A1 = 2 + 4 = 6

A2 = 2 - 4 = -2

out of

B = -4A + 9

we get

B1 = -4×6 + 9 = -24 + 9 = -15

B2 = -4×-2 + 9 = 8 + 9 = 17

now for slope PS and QR

(-11 - 1)/(A - 1) = (B - 5)/(5 - 2)

-12×3 = (B - 5)(A - 1) = AB - B - 5A + 5

-36 = AB - B - 5A + 5

0 = AB - B - 5A + 41

now using again the identity of the PQ slope ;

0 = A(-4A + 9) - (-4A + 9) - 5A + 41

0 = -4A² + 9A + 4A - 9 - 5A + 41 = -4A² + 8A + 32

0 = -A² + 2A + 8 = A² - 2A - 8

again solving the quadratic equation :

a = 1

b = -2

c = -8

x = A

A = (2 ± sqrt((-2)² - 4×1×-8))/(2×1) =

= (2 ± sqrt(4 + 32))/2 = (2 ± sqrt(36))/2 =

= (2 ± 6)/2 = 1 ± 3

A1 = 1 + 3 = 4

A2 = 1 - 3 = -2

and from

B = -4A + 9

we get

B1 = -4×4 + 9 = -16 + 9 = -7

B2 = -4×-2 + 9 = 8 + 9 = 17

so we get the possible solutions

1. P(6, -11), Q(5, -15)

2. P(-2, -11), Q(5, 17)

3. P(4, -11), Q(5, -7)

4. P(-2, -11), Q(5, 17)

the second criteria for a parallelogram is that the opposite sides must be equally long.

RS = sqrt((5 - 1)² + (2 - 1)²) = sqrt(16 + 1) = sqrt(17)

from the possible solutions we found solutions 2 and 4 are identical and their PQ side length would be

PQ = sqrt((17 - -11)² + (5 - -2)²) = sqrt(28² + 7²) = sqrt(833)

which is NOT sqrt(17).

so, these solutions are out.

but solutions 1 and 3 have the correct side length :

PQ = sqrt((-15 - -11)² + (5 - 6)²) = sqrt((-4)² + (-1)²) = sqrt(17)

and

PQ = sqrt((-7 - -11)² + (5 - 4)²) = sqrt(4² + 1²) = sqrt(17)

so, the correct values of A and B are

A = 6, B = -15

A = 4, B = -7

User Kumaraprasad
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