Sec 3 A-Math · Coordinate geometry

Mid-points and collinear points questions with answers

Practice questions on mid-points and collinear points, from the basics up to exam level, each with a full worked solution. Try them on paper first, then open the solution to check your working.

Worked example

Find the coordinates of the mid-point of the line joining $A(-3, 2)$ and $B(0, 7)$ .

Step 1 of 1

Mid-point = (average of the x's, average of the y's)

\mathstrut x = \dfrac{-3 + 0}{2}

x = -\dfrac{3}{2}

\mathstrut y = \dfrac{2 + 7}{2}

y = \dfrac{9}{2}

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Questions

Question 1The mid-point
Find the coordinates of the mid-point of the line joining $A(-3, 2)$ and $B(6, 3)$ .
Show the worked solution
1. Step 1: Mid-point = (average of the x's, average of the y's) $\mathstrut x = \dfrac{-3 + 6}{2}$ $x = \dfrac{3}{2}$ $\mathstrut y = \dfrac{2 + 3}{2}$ $y = \dfrac{5}{2}$
Question 2The mid-point
Find the coordinates of the mid-point of the line joining $A(4, -2)$ and $B(6, 3)$ .
Show the worked solution
1. Step 1: Mid-point = (average of the x's, average of the y's) $\mathstrut x = \dfrac{4 + 6}{2}$ $x = 5$ $\mathstrut y = \dfrac{-2 + 3}{2}$ $y = \dfrac{1}{2}$
Question 3The other end
The mid-point of $AB$ is $M(-3, 2)$ . Given that $A$ is $(5, 1)$ , find the coordinates of $B$ .
Show the worked solution
1. Step 1: M is the average of A and B: B = 2M − A $x = 2(-3) - 5$ $x = -11$ $y = 2(2) - 1$ $y = 3$
Question 4The other end
The mid-point of $AB$ is $M(5, 1)$ . Given that $A$ is $(6, 3)$ , find the coordinates of $B$ .
Show the worked solution
1. Step 1: M is the average of A and B: B = 2M − A $x = 2(5) - 6$ $x = 4$ $y = 2(1) - 3$ $y = -1$
Question 5Collinear points
The points $A(-3, 4)$ , $B(1, 2)$ and $C(9, k)$ lie on a straight line. Find the value of $k$ .
Show the worked solution
1. Step 1: Gradient of AB: put the coordinates into the gradient formula $m_{AB} = \dfrac{y_2 - y_1}{x_2 - x_1}$ $m_{AB} = \dfrac{2 - 4}{1 - (-3)}$ $= -\dfrac{1}{2}$
2. Step 2: A, B and C are on one line, so the gradient of AC is the same. Solve for k $\dfrac{k - 4}{9 - (-3)} = \dfrac{-1}{2}$ $k - 4 = -6$ $k = -2$
Question 6Collinear points
The points $A(1, 2)$ , $B(3, 6)$ and $C(5, k)$ lie on a straight line. Find the value of $k$ .
Show the worked solution
1. Step 1: Gradient of AB: put the coordinates into the gradient formula $m_{AB} = \dfrac{y_2 - y_1}{x_2 - x_1}$ $m_{AB} = \dfrac{6 - 2}{3 - 1}$ $= 2$
2. Step 2: A, B and C are on one line, so the gradient of AC is the same. Solve for k $\dfrac{k - 2}{5 - 1} = 2$ $k - 2 = 8$ $k = 10$

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Worked example

Find the coordinates of the mid-point of the line joining

A(-3, 2)

and

B(0, 7)

Step 1 of 1

Mid-point = (average of the x's, average of the y's)

\mathstrut x = \dfrac{-3 + 0}{2}

x = -\dfrac{3}{2}

\mathstrut y = \dfrac{2 + 7}{2}

y = \dfrac{9}{2}

Questions

Question 1The mid-point

Find the coordinates of the mid-point of the line joining $A(-3, 2)$ and $B(6, 3)$ .

Show the worked solution

Step 1: Mid-point = (average of the x's, average of the y's) $\mathstrut x = \dfrac{-3 + 6}{2}$ $x = \dfrac{3}{2}$ $\mathstrut y = \dfrac{2 + 3}{2}$ $y = \dfrac{5}{2}$

Question 2The mid-point

Find the coordinates of the mid-point of the line joining $A(4, -2)$ and $B(6, 3)$ .

Show the worked solution

Step 1: Mid-point = (average of the x's, average of the y's) $\mathstrut x = \dfrac{4 + 6}{2}$ $x = 5$ $\mathstrut y = \dfrac{-2 + 3}{2}$ $y = \dfrac{1}{2}$

Question 3The other end

The mid-point of $AB$ is $M(-3, 2)$ . Given that $A$ is $(5, 1)$ , find the coordinates of $B$ .

Show the worked solution

Step 1: M is the average of A and B: B = 2M − A $x = 2(-3) - 5$ $x = -11$ $y = 2(2) - 1$ $y = 3$

Question 4The other end

The mid-point of $AB$ is $M(5, 1)$ . Given that $A$ is $(6, 3)$ , find the coordinates of $B$ .

Show the worked solution

Step 1: M is the average of A and B: B = 2M − A $x = 2(5) - 6$ $x = 4$ $y = 2(1) - 3$ $y = -1$

Question 5Collinear points

The points $A(-3, 4)$ , $B(1, 2)$ and $C(9, k)$ lie on a straight line. Find the value of $k$ .

Show the worked solution

Step 1: Gradient of AB: put the coordinates into the gradient formula $m_{AB} = \dfrac{y_2 - y_1}{x_2 - x_1}$ $m_{AB} = \dfrac{2 - 4}{1 - (-3)}$ $= -\dfrac{1}{2}$
Step 2: A, B and C are on one line, so the gradient of AC is the same. Solve for k $\dfrac{k - 4}{9 - (-3)} = \dfrac{-1}{2}$ $k - 4 = -6$ $k = -2$

Question 6Collinear points

The points $A(1, 2)$ , $B(3, 6)$ and $C(5, k)$ lie on a straight line. Find the value of $k$ .

Show the worked solution

Step 1: Gradient of AB: put the coordinates into the gradient formula $m_{AB} = \dfrac{y_2 - y_1}{x_2 - x_1}$ $m_{AB} = \dfrac{6 - 2}{3 - 1}$ $= 2$
Step 2: A, B and C are on one line, so the gradient of AC is the same. Solve for k $\dfrac{k - 2}{5 - 1} = 2$ $k - 2 = 8$ $k = 10$