Question

what is the slope of a line perpendicular to the line whose equation is X+3y=-15

Answer 1

We know that :- product of slope of lines perpendicular to each other is = -1

For example if a be slope of line L1 and b be slope of line L2. Now if lines L1 and L2 are perpendicular to each other then :-

=> ab = -1

So, first we will find Slope of X+3y+15=0

=> slope of line of the form ax+by+c=0 is = -a/b

=> Slope of X+3y+15=0 is = -1/3

Let m be slope of line perpendicular to the line whose equation is X+3y=-15.

=> m × (-1/3) = -1

=> - m × (1/3) = -1

=> m/3 = 1

=> m = 1×3

=> m = 3

Answer :

• slope of a line perpendicular to the line whose equation is X+3y=-15 is 3

____________________

Additional Information :-

Co-ordinates of midpoint of line joining (l,m) and (v,u) is given as :-

$\rm \implies \bigg( \dfrac{l + v}{2} \: , \: \dfrac{m + u}{2} \bigg)$

Answer 2

We know that :- product of slope of lines perpendicular to each other is = -1

For example if a be slope of line L1 and b be slope of line L2. Now if lines L1 and L2 are perpendicular to each other then :-

=> ab = -1

So, first we will find Slope of X+3y+15=0

=> slope of line of the form ax+by+c=0 is = -a/b

=> Slope of X+3y+15=0 is = -1/3

Let m be slope of line perpendicular to the line whose equation is X+3y=-15.

=> m × (-1/3) = -1

=> - m × (1/3) = -1

=> m/3 = 1

=> m = 1×3

=> m = 3

Answer :

slope of a line perpendicular to the line whose equation is X+3y=-15 is 3

____________________

Additional Information :-

Co-ordinates of midpoint of line joining (l,m) and (v,u) is given as :-

$\rm \implies \bigg( \dfrac{l + v}{2} \: , \: \dfrac{m + u}{2} \bigg)$