JAMB Mathematics Past Questions & Answers - Page 423

\(y = x(2-x) \Rightarrow y= 2x - x^{2};

\int^{2}_{0}(2x-x^{2} = (x^{2}-\frac{x{3}}{3})^{2}\\
solving further gives (4 - \frac{1}{3} * 8) - (0) = \frac{4}{3} sq\hspace{1 mm}unit\)

The locus of a point P(x,y) such that PV = PW where V = (1,1) and W = (3,5). This means that the point P moves so that its distance from V and W are equidistance.

PV = PW

\(\sqrt{(x-1)^{2} + (y-1)^{2}} = \sqrt{(x-3)^{2} + (y-5)^{2}}\).

Squaring both sides of the equation,

(x-1)2 + (y-1)2 = (x-3)2 + (y-5)2.

x2-2x+1+y2-2y+1 = x2-6x+9+y2-10y+25

Collecting like terms and solving, x + 2y = 8.

No explanation has been provided for this answer.

Let \(\phi\) be the angle between the two lines.

tan \(\phi\) = \(\frac{m_1 - m_2}{1 + m_1 m_2}\)

where m\(_1\) = slope of line 1; m\(_2\) = slope of line 2.

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

Line 2: 3x - 2y = 5 \(\implies\) -2y = 5 - 3x.

y = \(\frac{3}{2}\)x - \(\frac{5}{2}\).

m\(_1\) = -2, m\(_2\) = \(\frac{3}{2}\).

tan \(\phi\) = \(\frac{-2 - \frac{3}{2}}{1 + (-2 \times \frac{3}{2})}\)

= \(\frac{\frac{-7}{2}}{-2}\)

\(\therefore\) Tan \(\phi\) = \(\frac{7}{4}\).

bisector theorem:

\(\frac{|MN|}{|MO|}\) = \(\frac{|PO|}{|NP|}\)

taking the bisected angle:x and y = |ON|=12

: x+y= 12

x =  12 - y

|PO| = 12 - y

\(\frac{6}{4}\)= \(\frac{12-y}{y}\)

6y = 4 (12-y)

6y = 48 - 4y

= 4.8

Recall that x+y= 12

12 - 4.8 =7.2