FINDING MISSING COORDINATE GIVEN SLOPE AND TWO POINTS

Find a given that the line joining :

Problem 1 :

A(1, 3) to B(3, a) is parallel to a line with gradient 3.

Solution :

If two lines are parallel, then

Given, A(1, 3) to B(3, a)

Slope m₁  =  (y₂ – y₁)/(x₂ – x₁)

  =  (a -3)/(3 – 1)

  =  (a – 3)/2  ---(1)

Slope of the given line (m₂)  =  3  ---(2)

m₁  =  m₂

(a – 3)/2  =  3

a – 3  =  6

a  =  9

Problem 2 :

P (a, -3) to Q(4, -2) is parallel to a line with gradient 1/3.

Solution :

Given, P (a, -3) to Q(4, -2)

=  (-2 + 3)/(4 – a)

m₁  =  1/(4 – a) ----(1)

m₂  =  1/3  ----(2)

m₁  =  m₂

1/(4 – a)  =  1/3

4 – a  =  3

 – a   =  3 – 4

a  =  1

Problem 3 :

M(3, a) to N(a, 5) is parallel to a line with gradient -2/5.

Solution :

Given, M(3, a) to N(a, 5)

m₁  =  (5 – a)/(a – 3)  ----(1)

m₁  =  m₂

m₂  =  -2/5 ----(2)

-2/5  =  (5 – a)/(a – 3)

-2(a – 3)  =  5(5 – a)

-2a + 6  =  25 - 5a

-2a + 6 – 25 + 5a  =  0

-19 + 3a  =  0

a  =  19/3

a  =  6 1/3

Find t given that the line joining :

Problem 4 :

A(2, -3) to B(-2, t) is perpendicular to a line with gradient 1 1/4.

Solution :

Since two lines are perpendicular then, 

m₁  x  m₂ =  -1

Given, A(2, -3) to B(-2, t)

m₁  =  (y₂ – y₁)/(x₂ – x₁)

=  (t + 3)/(-2 – 2)

=  (t + 3)/-4

m₁ x m₂  =  -1

(t + 3)/-4 x 1 1/4   =  -1

(t + 3)/-4  x  5/4  =  -1

(t + 3) x 5/-16  =  -1

(t + 3) x 5  =  16

5t + 15  =  16

5t  =  16 – 15

5t  =  1

t  =  1/5 

Problem 5 :

C(t, -2) to D(1, 4) is perpendicular to a line with gradient 2/3.

Solution :

Given, C(t, -2) to D(1, 4)

m₁  =  (y₂ – y₁)/(x₂ – x₁)

=  (4 + 2)/(1 – t)

=  6/(1 – t)

m₁ x m₂  =  -1

6/(1 – t)  x  2/3  =  -1

12/3(1 – t)  =  -1

12/3 – 3t  =  -1

12  =  -1 x(3 – 3t)

12  =  -3 + 3t

12 + 3  =  3t

15  =  3t

15/3  =  t

5  =  t

Problem 6 :

P(t, -2) to Q(5, t) is perpendicular to a line with gradient -1/4.

Solution :

Given, P(t, -2) to Q(5, t)

m₁  =  (y₂ – y₁)/(x₂ – x₁)

=  (t – (-2))/(5 – t)

=  (t + 2)/(5 – t)

m₁ x m₂  =  -1

(t + 2)/(5 – t) x -1/4  =  -1

(-t – 2)/(5 - t)/4 =  -1

(-t – 2)/(20 – 4t)  =  -1

-t – 2  =  -20 + 4t

-t – 2 + 20 – 4t  =  0

-5t + 18  =  0

t  =  18/5

t  =  3 3/5

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