How do you find the equation of the line t hat is perpendicular to x - 2y = -10 and passes through the point ( -1 , -3 )?

1 Answer
Sep 12, 2016

Answer:

#2x+y+5+0#.

Explanation:

Consider the following Results :

#"The eqn. of a line L' "bot" L : "ax+by+c=0" is given by"#

# L' : bx-ay+k=0, where, a^2+b^2ne0#.

# "In addition, L' passes through a pt. "(x_1,y_1)," then,"#

# L' : b(x-x_1)-a(y-y_1)=0#.

If we use these Results, we can immediately write the eqn. of reqd.

line as#2(x+1)+(y+3)=0, i.e., 2x+y+5+0#.

But, for the help of these Results, we can derive the eqn. as under :

Eqn. of the given line # : x-2y=-10, i.e., 2y=x+10, or, y=1/2x+5#.

So, the slope of this line is #1/2#, & hence, the slope reqd line (perp.

to given line) must be #-2#. We also have a pt.(-1,-3) on this line.

Hence, by the Slope-Pt. Form of line, the eqn. of reqd. line is,

#y+3=-2(x+1)#, which is the same as obtained earlier!.

Enjoy Maths.!