Given 2 pts P(a,b) and Q(c,d)

.

- The mid pt. coordinates of PQ are ((a+c)/2, (b+d)/2)

For example the mid pt. of P(1,3) and Q(−4,2) is ( (1−4)/2, (3+2)/2) = ( −3/2, 5/2)

- The distance of line PQ is given by PQ = √ 〈(c−a)² + (d−b)²〉

So in the above example, the distance PQ = √ 〈(1 + 4)² + (3 − 2)² 〉 = √ (25 + 1) = √ (26)

- The gradient of line PQ is given by, Gradient = tan θ = (d − b)/(c −a), whereby θ is the angle of the slope of line PQ make with the horizontal.

So in the above example, Gradient = (2 − 3)/(−4−1) = (−1)/(−5) = 1/5

Also if 0º < θ < 90º, the gradient is positive (tan θ is positive)

& if 90º < θ < 180º, the gradient is negative (tan θ is negative)

- Two lines are parallel is the gradients of the lines are same.
- Two lines are perpendicular to each other if the product of their gradients is −1. So if two lines have gradients of m and n respectively and the product mn = −1, we conclude that the two lines are perpendicular to each other.

*Example : Given 3 points A(2,3), B(5,4) and C(3,0). Show that AB is perpendicular to AC.*

*Let gradient AB = m = (4 − 3)/(5 − 2) = 1/3*

*and gradient AC = n = (3 − 0)/(2 − 3) = −3*

*product, mn = (1/3)(−3) = −1 ∴ AB is perpendicular to AC*

**Try these questions** :

1) The line 2x + y = 5 intersects the curve y² − 3x² = 20 − 14x at A and B. Find the length of AB and the coordinates of the mid point of AB.

2) The line joining the points A (4t, 4) and B (2t² − 3, 3) has gradient of 2. Find the possible values of t.

*Answers :*

1) y = 5 − 2x. substitute into the curve.

(5 − 2x)² − 3x² = 20 − 14x

25 − 20x + 4x² − 3x² + 14x − 20 = 0

x² − 6x + 5 = 0 ⇒ (x − 1)(x − 5) = 0 ⇒ x = 1 or 5.

when x = 1, y = 5 − 2(1) ∴ y = 3

when x = 5, y = 5 − 2(5) ∴ y = −5

So, A (1, 3) and B (5, −5)

∴ coordinates of the mid-pt. of AB = ( (1 + 5)/2, (3 − 5)/2) = **(3, −1)**

And the length of AB = √ 〈(−5 − 3)² + (5 − 1)²〉 = √ (64 + 16) = √80 =** 4√5**

2) Gradient of line AB = (3 −4)/(2t² − 3 − 4t) = 2

−1 = 2 (2t² − 3 − 4t)

4t² − 6 − 8t = −1 ⇒ 4t² − 8t − 5 = 0

(2t + 1)(2t −5) = 0 ⇒ ** t = −½ , 2½**

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