The equation of line passes (–1, 4) and perpendicular to the line 3x + 4y + 5 = 0 is About the author Ava
Answer: Equation of the given line: 3x + 4y -5 = 0 By rearranging, we get 4y = -3x + 5 or y = (-3/4)x + 5/4….(1) The standard equation of a straight line with slope m and y-intercept c is given by : y = mx + c….(2) Comparing (1) and (2), we get m = -3/4 and c = 5/4. Since the required line is perpendicular to (1), its slope M is -(1/m) = 4/3 We know that the required line passes through the point (5,1) The equation of this line is thus given by y-y1 =M(x-x1); where x1 is 5, y1 is 1 and M is 4/3 Substituting, we get y-1 = 4/3(x-5) Reply
Answer:
Equation of the given line: 3x + 4y -5 = 0
By rearranging, we get 4y = -3x + 5
or y = (-3/4)x + 5/4….(1)
The standard equation of a straight line with slope m and y-intercept c is given by :
y = mx + c….(2)
Comparing (1) and (2), we get m = -3/4 and c = 5/4.
Since the required line is perpendicular to (1), its slope M is -(1/m) = 4/3
We know that the required line passes through the point (5,1)
The equation of this line is thus given by y-y1 =M(x-x1); where x1 is 5, y1 is 1 and M is 4/3
Substituting, we get y-1 = 4/3(x-5)