Show the equation of a Line

The diagram shows part of the curve y = 2 − 18 2x + 3 , which crosses the x-axis at A and the y-axis at B. The normal to the curve at A crosses the y-axis at C. Show that the equation of the line AC is 9x + 4y = 27. Find the length of BC.

Answer ⤵ Crosses the x-axis means x-intercept when y = 0 y = 2 − 18 2x + 3 ⇒ when y = 0 0 = 2 − 18 2x + 3 0 = 4x + 6 − 18 2x + 3 0 = 4x − 12 12 = 4x 3 = x ⇔ A(3, 0) Crosses the y-axis means y-intercept when x = 0 y = 2 − 18 2x + 3 ⇒ when x = 0 y = 2 − 18 2·0 + 3 y = 2 − 6 y = − 4 ⇔ B(0, −4) The normal to the curve is Perpendicular (⊥) to it, we need the derivative to find its gradient. m1 = y' = 0 − 0(2x + 3) − 2(18) (2x + 3)² = 36 (2x + 3)² at the point A(3, 0), we have: m1 = 36 (2·3 + 3)² = 36 81 = 4 9 ⇔ m1 = 4 9 . AC ⊥ y ⇒ m1·m2 = − 1, m2 = − 1 m1 = − 1 4/9 = − 9 4 The gradient of the line AC is: m2 = − 9 4 Equation of the line AC Per definition the equation of a line is: y = mx + b m = m2, we need b. at the point A(3,0), 0 = − 9 4 · 3 + b ⇒ b = 27 4 Therefore, y = − 9 4 ·x + 27 4 4y = − 9x + 27 9x + 4y = 27 ⇒ True! The line crosses C when x = 0 (intercept y) 9·0 + 4y = 27 4y = 27 y = 27 4 ⇔ C(0, 27 4 ) The length of BC. d = √ [(xb + xc)² + (yb + yc)²] = √ [(0)² + ( − 4 + 27 4 )²] =√ ( 43 4 )² = 43 4      ⇒ BC = 10.75cm (or units)

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