- #1
Lytk
- 5
- 0
The co-ordinates of two points are A(4,-1) and B(7, -5)
The line L1: 3x+4y−12 = is perpendicular to AB.
(a) Find, in terms of k, the distance between the point P (10, k) and the line L2
(b) P(10, k) is on a bisector of the angles between the lines L1 and
L2 and L2 :5x+12y−20 =0
Find the possible values of k. I got part (a) using the formula to for the distance between a line and a point.
The answer was
$$ \frac{\left| 18-4k \right|}{5} $$
To find possible values of k , I used the same formula to find the distance between a line and the point and made it equal to the $$ \frac{\left| 18-4k \right|}{5} $$ , because its on the bisector
I got $$ \frac{\left| 30+12k \right|}{13} = \frac{\left| 18-4k \right|}{5} $$
$$ (5) \left| 30+12k \right|= (13) \left| 18-4k \right| $$
Theres only one step left I think but I'm just unsure of how to solve this
The line L1: 3x+4y−12 = is perpendicular to AB.
(a) Find, in terms of k, the distance between the point P (10, k) and the line L2
(b) P(10, k) is on a bisector of the angles between the lines L1 and
L2 and L2 :5x+12y−20 =0
Find the possible values of k. I got part (a) using the formula to for the distance between a line and a point.
The answer was
$$ \frac{\left| 18-4k \right|}{5} $$
To find possible values of k , I used the same formula to find the distance between a line and the point and made it equal to the $$ \frac{\left| 18-4k \right|}{5} $$ , because its on the bisector
I got $$ \frac{\left| 30+12k \right|}{13} = \frac{\left| 18-4k \right|}{5} $$
$$ (5) \left| 30+12k \right|= (13) \left| 18-4k \right| $$
Theres only one step left I think but I'm just unsure of how to solve this
Last edited: