41. Find the equation of a straight line which has slope -5/4 and passing through the point (-1,2).
Answers
Answer:
-5x-4y+3=0
Step-by-step explanation:
y-y1=m(x-x1)
y-2=-5/4(x-(-1))
y-2= -5/4 (x+1)
y-2= -5/4x -5/4
y=-5/4x-5/4+2
y = -5/4x +3/4
4y= -5x+3
-5x-4y+3=0
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Answer:
4y + 5x - 3 = 0
Step-by-step explanation:
Given that the straight line has a slope of -5/4 and it passes through the point (-1,2) we are asked to find the equation of straight line.
Whenever we are given one point and slope, we use slope point form of straight line which is given by,
- y - y1 = (x - x1)m
Here,
- x1 and y1 are the points on which the given line passes.
- m is the slope of line
So on substituting x1 = -1 and y1 = 2 and the value of slope = -5/4 in the slope point form of straight line, we get :
=> y - y1 = ( x - x1 ) m
=> y - 2 = ( x - (-1) ) (-5/4)
=> (y - 2)4 = ( x + 1 ) ( -5)
=> 4y - 8 = -5x - 5
=> 4y + 5x - 8 + 5 = 0
=> 4y + 5x - 3 = 0
Hence this is the required equation of straight line.