Question

The point A has coordinates (2, -5). The straight line 3x + 4y - 36 = 0 cuts the x-axis at B and cuts the y-axis at C. Find:
(a) the equation of the line through A which is perpendicular to the line BC.
(b) the perpendicular distance from A to the line BC.
(c) the area of triangle ABC.

Answer 1

I cant figure out c)

Answer 2

You can calculate the lengths of AB, BC and CA.

The area of the triangle A = sqrt( p(p-a)(p-b)(p-c) )
where p = (a+b+c)/2 where a,b,c are the three sides of the triangle.
This is called Heron's formula for the area of a triangle knowing the lengths of the 3 sides of the triangle.

Answer 3

Another way to find the area is: You know the equation of the line passing through BC. You know the equation of the line perpendicular to BC and passing though A. You can solve the two equations and find out where the two lines intersect. Call it D.
Then you can find the height of the triangle AD. Find BC. Area = 1/2 * BC ( AD.

Answer 4

herons law is not involved in our syllabus

Answer 5

I solved a) by y=-3/4x+9 so -1/(-3/4)=4/3

Answer 6

subsitute the gradient and the coordinates (2,-5)

Answer 7

giving us an equation 4x-3y=23

Answer 8

solve simultaneously between the equation 3x+4y=36, 4x-3y=23

Answer 9

giving us coordinates of (8,3)

Answer 10

I cant figure out c from this point

Answer 11

Call the point of intersection D. D is (8,3)
Find the distance AD. That will be the height of the triangle because AD is perpendicular to BC. Find BC. Area = 1/2 * base * height = 1/2 * BC * AD

Answer 12

AD is 10

Answer 13

i cant find the base

Answer 14

"The straight line 3x + 4y - 36 = 0 cuts the x-axis at B and cuts the y-axis at C."
Point B: set y = 0 to find x-intercept.
Point C: set x = 0 to find y-intercept.

Answer 15

((8-2)^2+(3--5)^2)^1/2

Answer 16

correct AD is 10.
Find the point B and C.

Answer 17

oh i found it thank you for your help

Answer 18

You are welcome.