Question 186716This question is from textbook mathematical analysis
: passing through (4,-5) and perpendicular to the line 3y= -2x/5 +3
please with full steps, Thankss
This question is from textbook mathematical analysis
Answer by uday1435(57)   (Show Source):
You can put this solution on YOUR website! You have given the condition but not the question. I presume you want the equation of a line passing through (4,- 5) and perpendicular to the line
3y = - (2/5)x +3
Ans: The condition for a line to be perpendicular to another line is that the product of their slopes must be - 1 . Means, one slope must be the negative reciprocal of the other
The given line is 3y = - (2/5) x + 3 . Rewriting it in y = mx+ b form we have
y = - (2/15) x + 1. So its slope = - (2/15) . The required line being perpendicular to this must have a slope = 15/2 and it passes through (4,- 5)
Let it be y = mx +b …. Eqn(1)
. when x = 4, y = -5 since it passes through (4, - 5). Plugging in these values ( m= 15/2, x = 4, y = - 5)into eqn (1) we get - 5 = (15/5) (4) + b
-5 = 12 + b ; so b = -5 -12 = -17 . Now we shall substitute the values of m and b in eqn(1)
y = (15/2) x – 17 which is the required equation. We can multiply it by 2 and make it into the standard form
2y = 15x – 17 ; ie. 15x – 2y – 17 = 0 is the required line.
If you still require further clarifications please write to udayakumar.t.r@gmail.com
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