13. Given that \( (a+b, a),(a-b, 2 a) \) and \( (b, c) \) are collinear points, express \( c \) in terms of \( a \) and \( b \).
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To determine \( c \) in terms of \( a \) and \( b \), we can use the concept of collinearity of points. Three points \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) are collinear if the area of the triangle they form is zero, which can be expressed as: \[ \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = 0. \] Substituting the coordinates of the given points: 1. \( (a+b, a) = (x_1, y_1) \) 2. \( (a-b, 2a) = (x_2, y_2) \) 3. \( (b, c) = (x_3, y_3) \) We have: \[ \begin{vmatrix} a+b & a & 1 \\ a-b & 2a & 1 \\ b & c & 1 \end{vmatrix} = 0. \] Calculating the determinant: \[ = (a+b)(2a - c) + (a)(c - 1) + (1)(b(2a) - (a-b)(a)). \] Let’s expand and simplify the determinant: 1. \( (a+b)(2a - c) = 2a^2 + 2ab - ac - bc \) 2. \( a(c - 1) = ac - a \) 3. \( b(2a) - (a-b)(a) = 2ab - (a^2 - ab) = 3ab - a^2 \) Now, we combine these: \[ 2a^2 + 2ab - ac - bc + ac - a + 3ab - a^2 = 0 \] This simplifies to: \[ a^2 + 5ab - bc - a = 0. \] Rearranging gives: \[ bc = a^2 + 5ab - a. \] Now, we solve for \( c \): \[ c = \frac{a^2 + 5ab - a}{b}. \] Thus, the expression for \( c \) in terms of \( a \) and \( b \) is: \[ c = \frac{a^2 + 5ab - a}{b}. \]