5. Triangle ABC has an angle of 90 degrees at B. Point A is on the y axis, AB is part of the line x-2y+8=0 and C is the point (6,2) (iii) Answer = AB equals the square root of 20 BC equals the square root of 20 Area of triangle equals 10 (iv) Using your answer to (iii), find the length of the perpendicular from B to AC. I don't even know where to start here, what is the method for figuring this out? And do they mean a line perpendicular to AC meeting B?

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5. Triangle ABC has an angle of 90 degrees at B. Point A is on the y axis, AB is part of the line x-2y+8=0 and C is the point (6,2) (iii) Answer = AB equals the square root of 20 BC equals the square root of 20 Area of triangle equals 10 (iv) Using your answer to (iii), find the length of the perpendicular from B to AC. I don't even know where to start here, what is the method for figuring this out? And do they mean a line perpendicular to AC meeting B?

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ok just making sure, the stuff you have for (iii) are distances you've already solved for?
Yes, and I checked the answers at the back of my book already so they are definitely correct.
Would it help to upload a picture of the triangle?

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yes definitely
Ok give me a second....
That is the triangle they have on the answers side of the book (the very first question was to draw it which i did successfully)
awesome :) ok. so it's saying the line from AC is \(x+3y-12=0\) so we need the line perpendicular AC that passes through B.
Did you solve for a point for B?
The point B is at (4,6) is that what you mean?
yes. there's a formula for this, the distance between a point and a line http://www.intmath.com/plane-analytic-geometry/perpendicular-distance-point-line.php But if you don't want to use that, we have to find the equation for the perpendicular line, then find the intersection point, then find the length of the segment
Ok, thats great and I could use that but would that be using the answers i got in the last question?
ok. I got you. Use the Pythagorean theorem to find the base of the triangle \[\sqrt{20}^2+\sqrt{20}^2=b^2\] Then use the formula for area of a triangle \[10=bh\] |dw:1441203786360:dw|
wow I'm a dummy. Area of a triangle is ½bh not just bh
you follow?
oh good I thought I was doing something wrong, yes let me just try it quickly to make sure I understand...
hmmmm I am getting square root of ten not just ten. My working out is: square root of 40 = b 10=1/2 x square root of 40 x h 10= square root of 10 x h square root of 10 = h
yes, √10 is correct
Oh sorry, have been making silly mistake with the maths all day, thank you so very much! You have helped me a lot!
you're welcome.

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