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Question 1152849: Find an equation of the straight line passing through the points with coordinates (4, −7)
and (−6, 11), giving your answer in the form ax + by + c = 0,where a, b and c are
integers.
The line crosses the x-axis at point A and the y-axis at point B and O is the origin.
(b) Find the area of triangle AOB.
Found 2 solutions by dkppathak, Alan3354:
Answer by dkppathak(439)   (Show Source):
You can put this solution on YOUR website! Find an equation of the straight line passing through the points with coordinates (4, −7)
and (−6, 11), giving your answer in the form ax + by + c = 0,where a, b and c are
integers.
The line crosses the x-axis at point A and the y-axis at point B and O is the origin.
(b) Find the area of triangle AOB.
slope =y2-y1/x2-x1 m=11+7/-6-4=18/-10=-9/5
equation of line
y+7=-9/5(x-4)
5y+35=-9x+36
9x+5y-1=0 point A on x axis (1/9,0) point B on Y axis (0,1/5) o point (0,0)
area of triangle AOB=1/2( basexalt)=1/2(1/45)=1/90
area =1/90 unit square
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Answer by Alan3354(69443)  (Show Source):
You can put this solution on YOUR website! Find an equation of the straight line passing through the points with coordinates (4, −7)
and (−6, 11), giving your answer in the form ax + by + c = 0,where a, b and c are
integers.
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Been done 1000's of times on this site.
I'll do 2 different points: (1,3) and (-3,5)
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Step 1, find the slope m.
m = diffy/diffx = (5-3)/(-3-1) = -1/2
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Use y-y1 = m*(x-x1) where (x1,y1) is either point.
y-3 = (-1/2)*(x-1)
Convert to any form you like.
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The line crosses the x-axis at point A and the y-axis at point B and O is the origin.
(b) Find the area of triangle AOB.
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Area = |A*B|/2 ----- |...| is absolute value
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