Using the Modulus to find a variable - to find co-ordinates

  • #1
Lytk
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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
 
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  • #2
Lytk said:
The co-ordinates of two points are A(4,-1) and B(7, -5)
The line L1: 3x+4y−12 = is perpendicular to AB.
Was this a "true false" question? If so the answer is "false". The slope of line AB is (-5- (-1))/(7- 4)= -4/3 and the slope of L1 is -4/3. The two lines are not] perpendicular since the product of the two slopes is not -1.

(a) Find, in terms of k, the distance between the point P (10, k) and the line L2
First you will have to define "L2"! Did you mean "L1" in the previous question or "L2" in the following question?

(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
 
  • #3
Hi ,
Sorry I didnt realize I made mistakes posting the question.

1) I had written the equation of the line wrong
It is 3x - 4y -12 =0 which makes the slope -3/-4 , which is 3/4
3/4 * -4/3 = -1

2) In (a) the question actually asked for the distance between P(10,k) and L1.
 

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