Question 1

LEVEL 6

Points A and B have coordinates (-6,11) and (0,8) respectively.

Point C lies on the line segment between A and B such that AC : CB = 1:2.

Find the coordinates of the point C.

Select the correct answer from the list below:

A: (-4,10)

B: (4,-10)

C: (0,0)

D: \left(-3, \frac{19}{2}\right)

 

CORRECT ANSWER:  A: (-4,10)

WORKED SOLUTION:

In a ratio of 1:2 there are 3 parts in total, and the distance from A to C constitutes 1 of those parts. Therefore, the distance from A to C counts for \frac{1}{3} of the total distance between A and B. So, we’re going to subtract the individual coordinates of A from B to find the distance in both x and y, and then we are going to add \frac{1}{3} of these respective distances to the coordinates of point A.

First, the x coordinates: 0-(-6)=6, then

\frac{1}{3}\times 6 = 2

Adding this to the x coordinate of A, gives the x coordinate of C:
- 6+2=-4

Second, the y coordinates: 8-11=-3, then

\frac{1}{3}\times(-3)=-1

Adding this to the y coordinate of A, gives the y coordinate of C:
11-1=10

Therefore, the coordinates of C are (-4,10)

Question 2

LEVEL 6

Points A and B have coordinates (-3,-2) and (6,4) respectively.

Point C lies on the line segment between A and B such that AC : CB = 3:2.

Find the coordinates of the point C.

Select the correct answer from the list below:

A: \left(\frac{12}{5}, \frac{8}{5}\right)

B:  \left(\frac{8}{5}, \frac{12}{5}\right)

C: (0,0)

D: \left(- \frac{42}{5}, - \frac{28}{5}\right)

 

CORRECT ANSWER:  A: \left(\frac{12}{5}, \frac{8}{5}\right)

WORKED SOLUTION:

In a ratio of 3:2 there are 5 parts in total, and the distance from A to C constitutes 1 of those parts. Therefore, the distance from A to C counts for \frac{3}{5} of the total distance between A and B. So, we’re going to subtract the individual coordinates of A from B to find the distance in both x and y, and then we are going to add \frac{3}{5} of these respective distances to the coordinates of point A.

First, the x coordinates: 6-(-3)=9, then

\frac{3}{5}\times 9 = \dfrac{27}{5}

Adding this to the x coordinate of A, gives the x coordinate of C:
- 3+\dfrac{27}{5}=\dfrac{12}{5}

Second, the y coordinates: 4-(-2)=6, then

\frac{3}{5}\times 6= \dfrac{18}{5}

Adding this to the y coordinate of A, gives the y coordinate of C:
-2 + \dfrac{18}{5}=\dfrac{8}{5}

Therefore, the coordinates of C are \left(\frac{12}{5}, \frac{8}{5}\right)

Question 3

LEVEL 6

The point C lies on the line segment AB such that AC : CB = 3:4.

Find the coordinates of C.

Select the correct answer from the list below:

A: \left(0.5,1\right)

B: \left(0,-2\right)

C: \left(\frac{2}{7},\frac{4}{7}\right)

D: \left(\frac{2}{7}, - \frac{8}{7}\right)

 

CORRECT ANSWER:   D: \left(\frac{2}{7}, - \frac{8}{7}\right)

WORKED SOLUTION:

In a ratio of 3:4 there are 7 parts in total, and the distance from A to C constitutes 3 of those parts. Therefore, the distance from A to C counts for \frac{3}{7}of the total distance between A and B. So, we’re going to subtract the individual coordinates of A from B to find the distance in both x and y, and then we are going to add \frac{3}{7} of these respective distances to the coordinates of point A.

First, the x coordinates: 2-(-1)=3, then

\frac{3}{7}\times3=\frac{9}{7}

Adding this to the x coordinate of A, gives the x coordinate of C:
-1+\frac{9}{7}=\frac{2}{7}

Second, the y coordinates: 4-(-5)=9, then

\frac{3}{7}\times9=\frac{27}{7}

Adding this to the y coordinate of A, gives the y coordinate of C:
-5+\frac{27}{7}= - \frac{8}{7}

Therefore, the coordinates of C are \left(\frac{2}{7}, - \frac{8}{7}\right)

Question 4

LEVEL 6

The point C lies on the line segment AB such that AC : CB = 5:1.

Find the coordinates of C.

Select the correct answer from the list below:

A: \left(\frac{7}{6},6\right)

B: \left(\frac{5}{6},\frac{6}{5}\right)

C: \left(-\frac{43}{6},2\right)

D: \left(\frac{7}{3},\frac{25}{6}\right)

 

CORRECT ANSWER:   A: \left(\frac{7}{6},6\right)

WORKED SOLUTION:

In a ratio of 5:1 there are 6 parts in total, and the distance from A to C constitutes 5 of those parts. Therefore, the distance from A to C counts for \frac{5}{6} of the total distance between A and B. So, we’re going to subtract the individual coordinates of A from B to find the distance in both x and y, and then we are going to add \frac{5}{6} of these respective distances to the coordinates of point A.

First, the x coordinates: 2-(-3)=5, then

\frac{5}{6}\times5=\frac{25}{6}

Adding this to the x coordinate of A, gives the x coordinate of C:
-3+\frac{25}{6}=\frac{7}{6}

Second, the y coordinates: 8-(-4)=12, then

\frac{5}{6}\times12=10

Adding this to the y coordinate of A, gives the y coordinate of C:
-4+10=6

Therefore, the coordinates of C are \left(\frac{7}{6},6\right)

Question 5

LEVEL 6

Points A and B have coordinates (-13,5) and (6,-13) respectively.

Point C lies on the line segment between A and B such that AC : CB = 7:11.

Find the coordinates of the point C.

Select the correct answer from the list below:

A: \left(\frac{11}{18},\frac{5}{2}\right)

B: \left(-\frac{101}{18},-2\right)

C: \left(1,1\right)

D: \left(\frac{19}{18},5\right)

 

CORRECT ANSWER:  B: \left(-\frac{101}{18},-2\right)

WORKED SOLUTION:

In a ratio of 7:11 there are 18 parts in total, and the distance from A to C constitutes 7 of those parts. Therefore, the distance from A to C counts for \frac{7}{18} of the total distance between A and B. So, we’re going to subtract the individual coordinates of A from B to find the distance in both x and y, and then we are going to add \frac{7}{18} of these respective distances to the coordinates of point A.

First, the x coordinates: 6-(-13)=19, then

\frac{7}{18}\times19=\frac{133}{18}

Adding this to the x coordinate of A, gives the x coordinate of C:
-13+\frac{133}{18}=-\frac{101}{18}

Second, the y coordinates: -13-5=-18, then

\frac{7}{18}\times(-18)=-7

Adding this to the y coordinate of A, gives the y coordinate of C:
5-7=-2

Therefore, the coordinates of C are \left(-\frac{101}{18},-2\right)