This is a topic which I haven't covered fully in college yet but need to know for an external college exam.

The vertices A,B,C of a triangle have position vector a,b,c. Point X lies on the line AB such that AX:XB = 5:2. Write an expression for the position vector of X in terms of a,b,c.

Find, in terms of a,b,c, the position vector of point P which divides XC in the ratio, XP:PC = y: 1-y

So far I have (2/7)a + (5/7)b = ->OX for the first bit

The vertices A,B,C of a triangle have position vector a,b,c. Point X lies on the line AB such that AX:XB = 5:2. Write an expression for the position vector of X in terms of a,b,c.

Find, in terms of a,b,c, the position vector of point P which divides XC in the ratio, XP:PC = y: 1-y

So far I have (2/7)a + (5/7)b = ->OX for the first bit

Original post by H_Fer

This is a topic which I haven't covered fully in college yet but need to know for an external college exam.

The vertices A,B,C of a triangle have position vector a,b,c. Point X lies on the line AB such that AX:XB = 5:2. Write an expression for the position vector of X in terms of a,b,c.

Find, in terms of a,b,c, the position vector of point P which divides XC in the ratio, XP:PC = y: 1-y

So far I have (2/7)a + (5/7)b = ->OX for the first bit

The vertices A,B,C of a triangle have position vector a,b,c. Point X lies on the line AB such that AX:XB = 5:2. Write an expression for the position vector of X in terms of a,b,c.

Find, in terms of a,b,c, the position vector of point P which divides XC in the ratio, XP:PC = y: 1-y

So far I have (2/7)a + (5/7)b = ->OX for the first bit

If you have OX then you can get the vector XC by following the sides of the triangle and then divide it in a similar manner to what you did originally, though the ratio is chosen to make the representation relatively simple.

If youve not done it, a sketch should help.

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