IGCSE Coordinate Geometry problem

Q. The straight line *l *passes through the points *A* and *B *with coordinates (2,0) and (5,6) respectively.**a)** Find an equation of *l*

The straight line *L* is perpendicular to *l* and passes through the point *C* with coordinates (0.5,7)

The lines *l* and *L* intersect in the point *D*.

**b) **Find an equation of *L* and show that the coordinates of *D* are (4.5,5)

Hence calculate

**c)** the ratio in which *D* divides the line segment *AB*

**d)** the area of triangle *ABC*

I've done all of it, but I'm having problems with part c. It would be really helpful if somebody showed me all the possible solutions to part c.

Please show all your working, and include the names of any facts or theories you use. If possible, include a diagram. Thank you! :)

Re: IGCSE Coordinate Geometry problem

It seems to be like its a really tough question why don't you take help from experts, you can go log on to www.artifextutors.com. You can chat with their live experts and take help from there.

Re: IGCSE Coordinate Geometry problem

I assure you, it's not a really tough question. It's ordinary level pure mathematics. Thanks for the advice though! :)

Re: IGCSE Coordinate Geometry problem

The question is not tough untill you loose confidence in solving the problem. Best of luck.

Re: IGCSE Coordinate Geometry problem

c) Find the distance from A to D and the distance from A to B. The ratio in which D divides AB is $\displaystyle \frac{|AD|}{|AB|}$

d) You have A=(2,0), B=(5,6), C=(0.5,7)

To find the area of the triangle they enclose first shift one of the points to the origin. A is the easiest. To move A from (2,0) to (0,0) you subtract 2 from the x coordinate and leave the y coordinate unchanged so to shift B and C the same just reduce their x coordinates by 2 and leave their y coordinates unchanged.

When you have the new positions of B and C, B' and C' you can find the area of the triangle with the formula

$\displaystyle Area= \frac{1}{2}|x_1y_2-x_2y_1| $

Where B'=(x_{1},y_{1}) and C'=(x_{2},y_{2})