Further Mathematics Questions and Answers

The point of intersection for the two lines exists at the point where the two lines are equal to each other. Make anyone of the variables the subject of the formula and equate the two lines to each other and solve for the coordinates of the point of intersection.

Given lines \(2y + 3x - 16 = 0\) and \(7y - 2x - 6 = 0\) , making x the subject of the formula:

Line 1 : \(2y + 3x - 16 = 0 \implies 3x = 16 - 2y\)

\(\therefore x = \frac{16}{3} - \frac{2}{3}y\)

Line 2 : \(7y - 2x - 6 = 0 \implies -2x = 6 - 7y\)

\(\therefore x = \frac{7}{2}y - 3\)

Equating them together and solving, we have:

\(\frac{16}{3} - \frac{2}{3}y = \frac{7}{2}y - 3  \implies \frac{16}{3} + 3 = \frac{7}{2}y + \frac{2}{3}y\)

\(\frac{25}{3} = \frac{25}{6}y \therefore y = 2\)

Putting y = 2 in the equation \(3x = 16 - 2y\), we have

\(3x = 16 - 2(2) = 16 - 4 = 12 \implies x = 4\)

The coordinate of P is (4, 2).

Given the endpoints of the diameter |EF|, the midpoint is the centre of the circle

= \((\frac{-2 + 3}{2} , \frac{-1 + 2}{2}) = (\frac{1}{2} , \frac{1}{2})\)

The radius is the distance from the centre to any point on the circle. Using \((\frac{1}{2}, \frac{1}{2})\) and \((3, 2)\);

\(r^{2} = (3 - \frac{1}{2})^{2} + (2 - \frac{1}{2})^{2} = \frac{25}{4} + \frac{9}{4}\)

\(r^{2} = \frac{34}{4}\)

The equation of a circle is given as:

\((x - a)^{2} + (y - b)^{2} = r^{2}\), (a, b) as the centre of the circle.

\(= (x - \frac{1}{2})^{2} + (y - \frac{1}{2})^{2} = \frac{34}{4}\)

\(x^{2} - x + \frac{1}{4} + y^{2} - y + \frac{1}{4} = \frac{17}{2}\)

= \(x^{2} - y^{2} - x - y - 8 = 0\) 

The gradient of the line y = 2x + 5 ;

\(\frac{\mathrm d y}{\mathrm d x} = 2\)

\(\frac{y - 3}{x - (-4)} = \frac{y-3}{x+4} = 2\)

\(y - 3 = 2(x + 4) \implies y = 2x + 11\)

\(\tan (x + y) = \frac{\tan x + \tan y}{1 - \tan x\tan y}\)

\(\tan x = \frac{5}{12} ; \tan y = \frac{3}{4}\)

\(\tan (x + y) = \frac{\frac{5}{12} + \frac{3}{4}}{1 - (\frac{5}{12} \times \frac{3}{4}})\)

= \(\frac{\frac{14}{12}}{\frac{33}{48}}\)

= \(\frac{56}{33}\)

\(\cos(a + b) = \cos a\cos b - \sin a\sin b\)

\(\cos75° = \cos(30 + 45) = (\cos30)(\cos45) - (\sin30)(\sin45)\)

= \((\frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2}) - (\frac{1}{2} \times \frac{\sqrt{2}}{2})\)

= \(\frac{\sqrt{6} - \sqrt{2}}{4}\)

= \(\frac{\sqrt{2}(\sqrt{3} - 1)}{4}\)