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

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The discussion revolves around finding the distance from a point P(10, k) to two lines, L1 and L2, defined by their equations. Initially, the poster calculated the distance to L2 incorrectly but later clarified that the distance should be to L1, yielding the formula |18 - 4k|/5. They also derived an equation to find possible values of k based on the point being on the angle bisector between L1 and L2, resulting in |30 + 12k|/13 = |18 - 4k|/5. A correction was made regarding the slope of L1, confirming that it is indeed perpendicular to line AB. The conversation highlights the importance of accurately defining line equations and understanding their geometric relationships.
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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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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
 
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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