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C1 Coordinate Geometry watch

1. Can someone help me with this question please......................

Part 1

Find an equation of the straight line passing through the points with coordinates (-1,5) and (4,-2),giving your answer in the form ax+by+c=0.

Part 2

The line crosses the x-axis at the point A and the y-axis at the point B,and O is the origin.

Find the area of triangle OAB?

Thanks your help would be much appreciated.............
2. 1. gradient = change in y/change in x = (5+2)/(-1-4)= -7/5

y-y1=(x-x1)dy/dx => y-5 = -7/5(x+1)

y= 3.6-7x/5
3. 2. When crosses x axis y=0

0=3.6-7x/5 => x=18/7

when crosses y axis x=0

y=3.6

area = r.a triangle = b*h/2 = 3.6*18/7 = 324/35
4. (Original post by vaider)
Can someone help me with this question please......................

Part 1

Find an equation of the straight line passing through the points with coordinates (-1,5) and (4,-2),giving your answer in the form ax+by+c=0.

Part 2

The line crosses the x-axis at the point A and the y-axis at the point B,and O is the origin.

Find the area of triangle OAB?

Thanks your help would be much appreciated.............
1) gradient = (-2 - 5) / (4 - -1)
= -7 / 5
- 1.4

equation of line:

y - y1 = m(x - x1)
y - 5 = -1.4x + 1.4
y = -1.4x +6.4
1.4x + y - 6.4 = 0

2) when crosses x-axis -> y = 0

1.4x - 6.4 = 0
x = 6.4/1.4

when crosses y-axis -> x = 0

y - 6.4 = 0
y = 6.4

A = (6.4/1.4 , 0)
B = (0, 6.4)
O = (0,0)

Area triangle = 1/2 base x height
= 3.2/1.4 x 6.4
= 14. 629 (3dp)

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