a. Let the numbers be p, r and q respectively Given: p + r + q = 81 --- (i) r = 2p --- (ii) q = r + 6 --- (iii) Substitute (2p) for r in equation (iii) q = 2p + 6 --- (iv) Substitute (2p) for r and (2p + 6) for q in equation (i) ⇒ p + (2p) + (2p + 6) = 81 ⇒ 5p + 6 = 81 ⇒ 5p = 81 - 6 ⇒ 5p = 75 p = \(\frac{75}{5}\) = 15. From equation (ii) ⇒ r = 2(15) = 30 From equation (iv) ⇒ q = 2(15) + 6 ⇒ q = 30 + 6 = 36 ∴ The numbers are 15, 30 and 36 b. \(m_1 = \frac{ y_2 - y_1}{ x_2 - x_1}\) Given points are P( 3, 5) and Q( -5, 7) = P(\( x_1 , y_1)\) , Q(\( x_2 , y_2)\) the slope of line PQ = \((\frac{ 7 - 5}{ -5 - 3}) = \frac{2}{-8} = -\frac{1}{4}\) Given that R( x, y) is the midpoint of the line formed by the points P( 3, 5) and Q( -5, 7) R(x, y) = \(\frac{3 + (-5)}{2}, \frac{5 + 7}{2} = \frac{ 3 - 5}{2} , \frac{ 5+7}{2}\) = -\(\frac{2}{2} , \frac{12}{2}\) = (-1, 6) but the line passes through the point R( -1, 6) and is perpendicular to LINE PQ \(m_2 = \frac{1}{m_1} = -\frac{1}{\frac{-1}{4}}\) = 4 Equation of the line can be obtained using \(y - y_1 = m_2( x - x_1)\) y - 6 = 4( x - (-1)) = y - 6 = 4( x + 1 ) y - 6 = 4x + 4 y = 4x + 4 + 6 Therefore, y = 4x + 10