Maths - coordinate geometry replied by popikachu @ Thu, 08 May 2008 21:06:00 +0800

popikachu — Thu, 08 May 2008 21:06:00 +0800

Answer:

y = mx +c

m = (3 - 2) / (4 - 1)
m = 1/3

y = (x/3) + c
2 = 1/3 + c
c = 5/3

Equation of line is y = (x/3) + 5/3

y = 0
0 = x/3 + 5/3
0 = x +5
x = -5

Coordinate of point (-5, 0)

Maths - coordinate geometry replied by popikachu @ Thu, 08 May 2008 21:01:23 +0800

popikachu — Thu, 08 May 2008 21:01:23 +0800

AC is a line...

A (1,2)
C (4,3)

If this line cut across the x-axis which means y = 0

First, find the equation of the line...
y = mx + c

then sub y in as 0
Find x.

Bingo, you have y and you have x now... thats the "coordinate of point" the question is asking ^^

16/f/lonely — Thu, 08 May 2008 20:51:12 +0800

That's simple. Find the gradient of AC.

The equation of AC would hence be y = gradient (x) + constant.

Sub any set of coordinates into the equation and you'll get constant.

The rest you should know.

bonkysleuth — Thu, 08 May 2008 20:15:36 +0800

the vertice of triangle ABC are A(01,2), B (1,5) and C(4,3).

a.Find the lengths of the sides AB, BC and CA

b. what type of triangle is ABC (RIGHT ANGLED ISOCELES)

c. find the perpendicular distance form B to ac.

d find the coordinates of the points at which the line AC cuts the x-axis

answer for a.

AB = square root 13 , same for BC and square root 26 for CA

(d) (-11,0)

please help with question d. thanks