WAEC 2019 · Paper 2 · Q2

(a)
Find the coordinates of the point which divides the line joining $(7, -5)$ and $(-2, 7)$ externally in the ratio $3 : 2$ .
Separate values with commas, e.g. 3, −2
(b)
Without using calculators or mathematical tables, evaluate $\dfrac{2}{1 - \sqrt2} - \dfrac{2}{2 + \sqrt2}$ , leaving the answer in the form $p + q\sqrt n$ , where $p$ , $q$ and $n$ are integers.

Worked solution (try it first)

(a)

For external division in the ratio $m : n = 3 : 2$ , use $\left(\dfrac{mx_2 - nx_1}{m - n}, \dfrac{my_2 - ny_1}{m - n}\right)$ , with $(x_1, y_1) = (7, -5)$ and $(x_2, y_2) = (-2, 7)$ .
The $x$ -coordinate: $\dfrac{3(-2) - 2(7)}{3 - 2} = -6 - 14$
$= -20$ .
The $y$ -coordinate: $\dfrac{3(7) - 2(-5)}{3 - 2} = 21 + 10$
$= 31$ .
The point is $(-20, 31)$ .

(b)

Put the two fractions over the common denominator $(1 - \sqrt2)(2 + \sqrt2)$ : the top is $2(2 + \sqrt2) - 2(1 - \sqrt2) = 2 + 4\sqrt2$ .
The bottom is $(1 - \sqrt2)(2 + \sqrt2) = 2 + \sqrt2 - 2\sqrt2 - 2$
$= -\sqrt2$ .
So the value is $\dfrac{2 + 4\sqrt2}{-\sqrt2}$ .
Multiply top and bottom by $\sqrt2$ : $\dfrac{2\sqrt2 + 8}{-2} = -4 - \sqrt2$ .

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