Math, asked by priyadharshinisumith, 3 months ago

41. Find the equation of a straight line which has slope -5/4 and passing through the point (-1,2).​

Answers

Answered by SaavniGupta
3

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

Hope it helps! please mark me brainliast!

Answered by Anonymous
1

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.

Similar questions