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1)calculate the area of the triangle formed by the line 3x + 4y = 24 and the axes.

2) sketch of the line with the point (2,6) marked on it and also the point (4,9). Find the shortest distance between the point (4,9 ) and the line 3y+2x=22

Please could you show detailed working out so i understand how to do it next time.
2. I got:

1) 24

2) √13

Are they right?
3. (Original post by AlexOD)
I got:

1) 24

2) √13

Are they right?
Yes, according to the mark scheme in the book both of those are correct but I don't know how to work it out.
4. So basically, for question 1, you first need to rearrange the formula you are given for the line. This rearranges to: y=-3/4(x)+6 So from this equation you can draw on where the line meets the axes. So go ahead and sub 0 in for x and then for y. You should get 2 coordinates: (8,0) and (0,6) You'll notice, if you draw the line through both of these points, that you get a right-angled triangle made of the points (0,0), (8,0) and (0,6). You should hopefully find the last part quite easy, as you can see, the lengths of either side of the triangle are 6 and 8. So for the last step, just do (6*8)/2 Thus, the area = 24.
5. For question 2, it appears that they have given you the closest point to (4,9), as I find it to be the point (2,6). I did manage to work that out too so I could provide a method for finding the closest point too if you'd like? But, all you have to do on this question is use the Pythagorean theorem between the two points. See, to find the distance between any point you need to make a triangle where one point is at the bottom of the hypotenuse and the other point is at the top. With this you find that the length of the bottom of the triangle (being the distance between the x coordinates, 4 and 2) is 2, and the distance between the side of the triangle (being the distance between the y coordinates i.e 9-6) is 3. Therefore, √(3^2+2^2) = √13.
6. Anywhere you feel I need to elaborate on?
7. (Original post by AlexOD)
So basically, for question 1, you first need to rearrange the formula you are given for the line. This rearranges to: y=-3/4(x)+6 So from this equation you can draw on where the line meets the axes. So go ahead and sub 0 in for x and then for y. You should get 2 coordinates: (8,0) and (0,6) You'll notice, if you draw the line through both of these points, that you get a right-angled triangle made of the points (0,0), (8,0) and (0,6). You should hopefully find the last part quite easy, as you can see, the lengths of either side of the triangle are 6 and 8. So for the last step, just do (6*8)/2 Thus, the area = 24.
Thank you so much, i finally understand it.
8. Glad to hear it my friend Coordinate geometry will be a breeze come january/february time trust me.
9. (Original post by AlexOD)
For question 2, it appears that they have given you the closest point to (4,9), as I find it to be the point (2,6). I did manage to work that out too so I could provide a method for finding the closest point too if you'd like? But, all you have to do on this question is use the Pythagorean theorem between the two points. See, to find the distance between any point you need to make a triangle where one point is at the bottom of the hypotenuse and the other point is at the top. With this you find that the length of the bottom of the triangle (being the distance between the x coordinates, 4 and 2) is 2, and the distance between the side of the triangle (being the distance between the y coordinates i.e 9-6) is 3. Therefore, √(3^2+2^2) = √13.
Thank you, I don't know what I'd do without your help. ☺
10. Welcome
11. (Original post by AlexOD)
Glad to hear it my friend Coordinate geometry will be a breeze come january/february time trust me.
I hope so. we received a sheet of 75 questions on things like this and i understood most except a few. If you have time and don't mind helping me there's one last one I'd appreciate help on.

Point A (2,6) is perpendicular to PQ whose equation is 2y-x=5. Find the length of the perpendicular from A to PQ.
I already worked out the equation of the line A perpendicular to PQ as y+2x=10. And i know i need to use the formula for length of the line. But I'm not sure what my x2 and y2 values are.
12. Ok, so basically your x2 and y2 coordinates are where the line PQ intersects the perpendicular line through A. So, by making 'y' the subject of your equation for the perpendicular A and subbing it into the equation for the line PQ, you can find the x coordinate. Subbing this x coordinate into any one of the equations will suffice in order to find the corresponding y coordinate.
13. (Original post by AlexOD)
Ok, so basically your x2 and y2 coordinates are where the line PQ intersects the perpendicular line through A. So, by making 'y' the subject of your equation for the perpendicular A and subbing it into the equation for the line PQ, you can find the x coordinate. Subbing this x coordinate into any one of the equations will suffice in order to find the corresponding y coordinate.
Thank you, I finally got the answer. Before i tried to solve them by simultaneous equation but it wouldn't give me he right answer but i got it now, so thank you and good night.
14. You too buddy

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