Question

find the equation of the line perpendicular to the linex -2y+3=0 and passing through the point(1,-2) ??????

Answer 1

✨✨✨✨ Heya ✨✨✨✨

Given,

line x -2y+3=0

➡
$y = \frac{1}{2} x + \frac{3}{2}$
Slope of line is M1 .

•°• , the slope of the line perpendicular to line one is..

$M_2 =$ $- \frac{1}{m1} = - 2$

✨equation of the line with slope -2 and passing through the point( 1 ,- 2 )is

y -(-2) = -2(x-1) or

y+ 2x= 0
so you get the answer..y +2x =0

Answer 2

HEYA MATE!
HERE'S YOUR ANSWER!

x-2y+3=0
=>2y=x+3
=>y=(1/2)x + (3/2)
So, slope (m1) is 1/2
Now , the two lines are perpendicular.
Let slope of 2nd line be m2
So, m2=(-1)/m1 =-2
So, equation of the line passing through (1,-2) having slope =-2 is,
(y-y1) = m (x-x1)
=> y-(-2) = m2 (x-1)
=> y+2 = -2(x-1)
=> y+2 = -2x+2
=> 2x+y=0

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