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
(C) 2.55 units
(d) (-11,0)
please help with question d. thanks
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.
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 ^^
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)