Are the points M (−3, −4), A (0, 0) and G (6, 8) on the same straight line? Suggest a method to check this without plotting and joining the points.
Step-by-step solution
To find: Whether M, A, G are collinear, and a method that needs no drawing
Idea: If three points are on one line, the middle one splits the longest distance into the other two: the two shorter distances add up exactly to the longest. If they form a triangle, the sum of two sides is always greater than the third.
- Method: find the distance between each pair of points with the distance formula. If the two smaller distances add up to the largest, the points are on one straight line; if not, they form a triangle.½ mark
- MA = √((0 − (−3))2 + (0 − (−4))2) = √(9 + 16) = √25 = 5.½ mark
- AG = √((6 − 0)2 + (8 − 0)2) = √(36 + 64) = √100 = 10.½ mark
- MG = √((6 − (−3))2 + (8 − (−4))2) = √(81 + 144) = √225 = 15.½ mark
- MA + AG = 5 + 10 = 15 = MG. So A lies on the segment MG, and M, A and G are on the same straight line.1 mark
- (A second way to see it: from M to A the shift is 3 across and 4 up; from A to G it is 6 across and 8 up, exactly twice as much in the same direction. So the path does not turn at A.)
Check: Every point on the line through O and (3, 4) has y = 43x: for M, 43 × (−3) = −4 ✓; for G, 43 × 6 = 8 ✓.
Answer to write in the exam
Method: if the sum of the two smaller distances equals the largest, the points are collinear.
MA = √((0 + 3)2 + (0 + 4)2) = √25 = 5
AG = √(62 + 82) = √100 = 10
MG = √((6 + 3)2 + (8 + 4)2) = √225 = 15
MA + AG = 5 + 10 = 15 = MG
∴ M, A, G lie on the same straight line (A between M and G).
Common mistakes that cost marks
- Checking only two distances. Two points are always on a line; you need all three distances to compare.
- Adding the wrong pair: the two smaller distances must add up to the largest.
- Sign slips such as 0 − (−3) = −3. Subtracting a negative adds: 0 + 3 = 3.
How this can come in the exam
For three points P, Q, R, PQ = 4, QR = 7 and PR = 11. Then
- P, Q, R are collinear with Q between P and R
- PQR is a right-angled triangle
- PQR is an isosceles triangle
- R lies between P and Q
Show answer
(A) P, Q, R are collinear with Q between P and R
PQ + QR = 4 + 7 = 11 = PR, so Q lies on segment PR.
Check whether the points (−1, 3), (1, 7) and (4, 13) lie on one straight line.
Show answer
Distances √(22 + 42) = √20 = 2√5 (½ mark), √(32 + 62) = √45 = 3√5 (½ mark), √(52 + 102) = √125 = 5√5 (1 mark). 2√5 + 3√5 = 5√5, so yes, they are collinear (1 mark).Try one yourself
Are the points (−2, −3), (−1, 1) and (2, 13) on one straight line?
Show answer
Distances √(1 + 16) = √17, √(9 + 144) = √153 = 3√17 and √(16 + 256) = √272 = 4√17. Since √17 + 3√17 = 4√17, yes, they are collinear.
More questions like this
- Use your method (the one used for M (−3, −4), A (0, 0) and G (6, 8)) to check if the points R (−5, −1), B (−2, −5) and C (4, −12) are on the same straight line. Now plot both sets of points and check your answers.
- Using the origin as one vertex, plot the vertices of:
- The following table shows the coordinates of points S, M and T. In each case, state whether M is the midpoint of segment ST. Justify your answer.
- Use the connection you found to find the coordinates of B given that M (−7, 1) is the midpoint of A (3, −4) and B (x, y).
- Let P, Q be points of trisection of AB, with P closer to A, and Q closer to B. Using your knowledge of how to find the coordinates of the midpoint of a segment, how would you find the coordinates of P and Q? Do this for the case when the points are A (4, 7) and B (16, −2).