# Points A(0,5) B(9,8) C(4,3). Find a vector condition equation for the line,p, which passes through the point C and which is perpendicular to the line AB.

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The perpendicularity of two vectors is expressed by the the dot product of the vectors.

If this dot product cancels, then the vectors are perpendicular to each other.

You need to find the vector of the line AB:

`bar (r) =bar (r_1) + s(bar (r_2) - bar (r_1))`

`` `bar (r_1) = (0)* bar i + (5)*bar j`

`` `bar (r_2) = (9)* bar i + (8)*bar j`

`` `bar (r) =(5)*bar j + s((9)* bar i + (8)*bar j -(5)*bar j)`

`bar (r) =(5)*bar j + s(9 bar i + 3 bar j)`

The vector that gives the perpendicular that passes through the point C and it falls to the line AB is `bar (CP).`

`bar (CP) = (x_P - x_C)* bar i + (y_P - y_C)*bar j`

`bar (CP) = (x_P - 4)* bar i + (y_P - 3)*bar j`

You know that `bar (AB)*bar (CP)` = 0

`(9*bar i + 3*bar j)*``((x_P - 4)* bar i + (y_P - 3)*bar j)` = 0

`9s(x_P - 4) + (5+3s)(y_P - 3) = 0`

**Hence, the vector that passes through C and it is perpendicular to the director vector of AB respects the rule** `(9*bar i + 3*bar j)((x_P - 4)* bar i + (y_P - 3)*bar j)=0`