Question

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

Answer 1

`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.!