Help please (1 Viewer)

Miz Janani · Oct 2, 2009

Sample HSC Papers : Mathematics

Im doing past hsc papers as a means of practicing for the hsc. Could someone please advice me how to do question 2bv.

Thankyou in advanced.

cyclicpermutate · Oct 2, 2009

Miz Janani said:
Sample HSC Papers : Mathematics

Im doing past hsc papers as a means of practicing for the hsc. Could someone please advice me how to do question 2bv.

Thankyou in advanced.

So there are three parts to this question. Firstly you must find the point C, then find the equation of the line BC, then use the perpendicular distance formula. Note that it says that ABCO is a parallelogram. This means that AO is the 'same' as BC.

To get from A to O you must go 2 units to the right and then 5 units down. So we shall do the same thing for B to C. The point B is (4,3), so we must go across 2 on the x axis (this is the same as adding 2 to the x coordinate), then we must go down 5 (this is the same as subtracting 5 from the y coordinate).

Therefore C is (4+2, 3-5) ie (6, -2). Now we must use the 2 point formula. Note that we do this because we have two points so its easier.

$\frac{y-y_{1}}{x-x_{1}} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$

$\frac{y - 3}{x - 4} = \frac{-2 - 3}{6 - 4}$

$\frac{y - 3}{x - 4} = \frac{-5}{2}$

$2\left(y - 3\right) = -5 \left(x - 4\right)$

$2y - 6 = -5x + 20$

$5x + 2y - 26 = 0$

then we use the perpendicular distance formula

$d = \frac{|ax_{1} + by_{1} + c|}{\sqrt{a^2 + b^2}}$

$= \frac{|5 \times 0 + 2 \times - 26|}{\sqrt{5^2 + 2^2}}$

$= \frac{26}{\sqrt{29}}$

$= \frac{26\sqrt{29}}{29}$

Help please (1 Viewer)

Miz Janani

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cyclicpermutate

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