Area from a Line Equation

[heleep]

"Find the equation of the straight line through A(1,4) which is perpendicular to the line passing through the points B(2,-2) and C(4,0).
Hence find the area of the triangle ABC, giving your answer in the simplest possible form."


Now the equation for for the perpendicular line through point A is y=-x+5, but what relevance has this got in finding the area of the triangle?
To find the area, I thought I had to use coordinate geometry to determine the lengths.
How do I use the equation? The answer says it's 7?

 

[John Marsh]

Given the straight line through A(1,4) which is perpendicular to the line passing through the points B(2,-2) and C(4,0)

First find the slope
[0-(-2)]/(4-2)=1

As we know the equation
y=mx+b
b=y-mx
b=-2-1*2
b=-4
y=x-4

Now find the length BC
√[(2-4)^2+(-2-0)^2]
√8
2√2

Line containing A and perpendicular to line BC
Want slope -1.

Line
y-4=-x+1*2y=-x-3

Point of intersection these two lines
x-4=-x-3
2x-4=-3
2x=4-3
2x=1
x=1/2

Either equation line, y=-x-3
y=-1/2-6/2
y=7/2

Point on line BC containing A is (1/2,-7/2)

Find the distance from A(1,4) to (1/2, -7/2), call the distance AP.

Find the area of the triangle ABC using
1/2*BC*AP
​

 

[heleep]

I see. I was thinking 'why can't I just use the midpoint of B and C', but I wasn't thinking.

You lost me at:
"Line
y-4=-x+1*2y=-x-3"

Where did that come from?
It doesn't matter. I've gotten the right answer now, thanks.

 

[HallsofIvy]

"Find the equation of the straight line through A(1,4) which is perpendicular to the line passing through the points B(2,-2) and C(4,0).
Hence find the area of the triangle ABC, giving your answer in the simplest possible form."


Now the equation for for the perpendicular line through point A is y=-x+5, but what relevance has this got in finding the area of the triangle?
To find the area, I thought I had to use coordinate geometry to determine the lengths.
How do I use the equation? The answer says it's 7?

The line from B(2, -2) to C(4, 0) has slope \(\displaystyle \frac{0- (-2)}{4- 2}= \frac{2}{2}= 1\) so, yes, a line "perpendicular to the line passing through the points B(2,-2) and C(4,0)" has slope -1. It must be of the form y= -x+ b. Since it passes through A(1, 4) we must have 4= -1+ b so b= 5. Yes, that line is y=-x+ 5. You want that in order to find the point where A crosses BC. That happens when both equations are satisfied. The line from B to C has, as before, slope 1 so is of the form y= x+ b. The fact that it passes through B(2, -2) means that -2= 2+ b so b = -4. The equation of the line is y= x- 4. (As a check, x= 4, y= 0 also satisfies that equation.

At the point of intersection, y= -x+ 5= x- 4 so that 2x= 9, x= 9/2 and y= -9/2+ 5= 9/2- 4= 1/2. The point of intersection is (9/2, 1/2).
The area of a triangle is "1/2 base times height". The "base" is the distance from B to C and the "height" is the distance from A to (9/2, 1/2).