The coordinates of points A and B are (-5,6) and (4, – 6) respectively. The point C lies on the line AB, between A and B, such that AC/CB = 1/2.
- (a)Find the coordinates of C.[2]
- (b)The line CD is perpendicular to AB. Find the equation of CD in the form y = mx+c.[4]
- (c)The length of BD is √125. Find the coordinates of the two possible positions of point D.[6]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2022). The mark or mark type is shown in brackets.
- [B2] (-2, 2) (B1 for one correct coordinate nfww)
- [M1] for AC = 1/3 AB or 1/3 (-9 -12)
- [M1] or CA = 1/3 BA or 1/3 (9 12)
- [M1] for x = -5 + 3 or x = 4 - 6 or for y = 6 - 4 or y = -6 + 8
- [M1] or for x+5/2 = (4-x)/2 or (6-y)/2 = (y+6)/2
- [M1] or for 2OC - OA = OB - OC oe
- [B1] m_AB = (-6-6)/(4-(-5)) oe or -12/9 or -4/3
- [M1] m_CD = 3/4 FT their m_AB
- [M1] y - 2 = (3/4)(x + 2) oe FT their (-2, 2) and -1/their m_AB
- [M1] or y = (3/4)x + c and 2 = (3/4)(-2) + c oe soi
- [A1] y = (3/4)x + 7/2 or equivalent in form y = mx+c
- [B1] (x-4)² + (y+6)² = 125 oe, soi
- [M1] Uses their y = (3/4)x + 7/2 to eliminate one unknown (if correct implies B1)
- [A1] Correct equation in one unknown: (x-4)² + ((3/4)x + 7/2 + 6)² = 125 oe
- [A1] Writes in solvable form: 25x² + 100x - 300 = 0 oe
- [A1] Factorises or solves a correct 3-term quadratic
- [A1] (2, 5) and (-6, -1)
- [SC2] for identifying one correct point by inspection from the length equation and testing it in the correct equation of CD
- [SC2] for identifying the second correct point by inspection from the length equation and testing it in the correct equation of CD
- [B1] [BC =] sqrt((4-(their-2))^2 + (-6-(their2))^2) (Alternative method)
- [M1] CD = sqrt(125 - their BC^2) providing 125 - their 100 > 0 (Alternative method)
- [A1] CD = 5 (Alternative method)
- [A2] (x-C_x)^2 + (y-C_y)^2 = 5^2 (A1 for CD_x or CD_y) (Alternative method)
- [A1] (2, 5) and (-6, -1) (Alternative method)
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