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Answer :
Let's solve this problem step by step:
### (i) Distance from the point [tex]\( P(10, k) \)[/tex] to the line [tex]\( l_1: 3x - 4y - 12 = 0 \)[/tex]
The formula to find the distance from a point [tex]\((x_1, y_1)\)[/tex] to a line [tex]\(ax + by + c = 0\)[/tex] is:
[tex]\[
\text{Distance} = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}
\][/tex]
For the line [tex]\( l_1: 3x - 4y - 12 = 0 \)[/tex], we have:
- [tex]\(a = 3\)[/tex],
- [tex]\(b = -4\)[/tex],
- [tex]\(c = -12\)[/tex].
The point is [tex]\( P(10, k) \)[/tex], so [tex]\( x_1 = 10 \)[/tex] and [tex]\( y_1 = k \)[/tex].
Applying the formula:
[tex]\[
\text{Distance} = \frac{|3(10) - 4(k) - 12|}{\sqrt{3^2 + (-4)^2}}
\][/tex]
[tex]\[
= \frac{|30 - 4k - 12|}{\sqrt{9 + 16}}
\][/tex]
[tex]\[
= \frac{|18 - 4k|}{5}
\][/tex]
So, the distance is [tex]\( \frac{|18 - 4k|}{5} \)[/tex].
### (ii) Point [tex]\( P(10, k) \)[/tex] is on the bisector between lines [tex]\( l_1 \)[/tex] and [tex]\( l_2: 5x + 12y - 20 = 0 \)[/tex]
For a point to lie on the angle bisector of two lines, the point must satisfy the condition that the distance to both lines (in absolute terms) is the same.
Using the distance formula again, let's calculate the distance to line [tex]\( l_2: 5x + 12y - 20 = 0 \)[/tex].
- For this line [tex]\( a_2 = 5 \)[/tex], [tex]\( b_2 = 12 \)[/tex], [tex]\( c_2 = -20 \)[/tex].
The distance from [tex]\( P(10, k) \)[/tex] to [tex]\( l_2 \)[/tex] is:
[tex]\[
\text{Distance to } l_2 = \frac{|5(10) + 12(k) - 20|}{\sqrt{5^2 + 12^2}}
\][/tex]
[tex]\[
= \frac{|50 + 12k - 20|}{\sqrt{25 + 144}}
\][/tex]
[tex]\[
= \frac{|30 + 12k|}{13}
\][/tex]
Since [tex]\( P(10, k) \)[/tex] is on the bisector, the distances are equal:
[tex]\[
\frac{|18 - 4k|}{5} = \frac{|30 + 12k|}{13}
\][/tex]
To solve for [tex]\( k \)[/tex], cross-multiply and simplify:
[tex]\[
13|18 - 4k| = 5|30 + 12k|
\][/tex]
Now solve these equations separately for [tex]\( k \)[/tex]:
1. [tex]\( 13(18 - 4k) = 5(30 + 12k) \)[/tex]
2. [tex]\( 13(18 - 4k) = -5(30 + 12k) \)[/tex]
Solution of 1:
[tex]\[
234 - 52k = 150 + 60k
\][/tex]
[tex]\[
234 - 150 = 60k + 52k
\][/tex]
[tex]\[
84 = 112k
\][/tex]
[tex]\[
k = \frac{84}{112} = \frac{3}{4}
\][/tex]
Solution of 2:
[tex]\[
234 - 52k = -150 - 60k
\][/tex]
[tex]\[
234 + 150 = 52k - 60k
\][/tex]
[tex]\[
384 = -8k
\][/tex]
[tex]\[
k = -\frac{384}{8} = -48
\][/tex]
The possible values of [tex]\( k \)[/tex] are [tex]\( \frac{3}{4} \)[/tex] and [tex]\(-48\)[/tex].
### (ii) (a)
The possible values of [tex]\( k \)[/tex] are [tex]\( k = \frac{3}{4} \)[/tex] and [tex]\( k = -48 \)[/tex].
### (ii) (b) If [tex]\( k > 0 \)[/tex], find the distance from [tex]\( P \)[/tex] to [tex]\( l_1 \)[/tex].
Since [tex]\( k = \frac{3}{4} \)[/tex] is the only positive value:
Substitute [tex]\( k = \frac{3}{4} \)[/tex] into the distance formula for [tex]\( l_1 \)[/tex]:
[tex]\[
\text{Distance} = \frac{|18 - 4(\frac{3}{4})|}{5}
\][/tex]
[tex]\[
= \frac{|18 - 3|}{5}
\][/tex]
[tex]\[
= \frac{15}{5}
\][/tex]
[tex]\[
= 3
\][/tex]
So, the distance from [tex]\( P \)[/tex] to [tex]\( l_1 \)[/tex] when [tex]\( k > 0 \)[/tex] is 3.
### (i) Distance from the point [tex]\( P(10, k) \)[/tex] to the line [tex]\( l_1: 3x - 4y - 12 = 0 \)[/tex]
The formula to find the distance from a point [tex]\((x_1, y_1)\)[/tex] to a line [tex]\(ax + by + c = 0\)[/tex] is:
[tex]\[
\text{Distance} = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}
\][/tex]
For the line [tex]\( l_1: 3x - 4y - 12 = 0 \)[/tex], we have:
- [tex]\(a = 3\)[/tex],
- [tex]\(b = -4\)[/tex],
- [tex]\(c = -12\)[/tex].
The point is [tex]\( P(10, k) \)[/tex], so [tex]\( x_1 = 10 \)[/tex] and [tex]\( y_1 = k \)[/tex].
Applying the formula:
[tex]\[
\text{Distance} = \frac{|3(10) - 4(k) - 12|}{\sqrt{3^2 + (-4)^2}}
\][/tex]
[tex]\[
= \frac{|30 - 4k - 12|}{\sqrt{9 + 16}}
\][/tex]
[tex]\[
= \frac{|18 - 4k|}{5}
\][/tex]
So, the distance is [tex]\( \frac{|18 - 4k|}{5} \)[/tex].
### (ii) Point [tex]\( P(10, k) \)[/tex] is on the bisector between lines [tex]\( l_1 \)[/tex] and [tex]\( l_2: 5x + 12y - 20 = 0 \)[/tex]
For a point to lie on the angle bisector of two lines, the point must satisfy the condition that the distance to both lines (in absolute terms) is the same.
Using the distance formula again, let's calculate the distance to line [tex]\( l_2: 5x + 12y - 20 = 0 \)[/tex].
- For this line [tex]\( a_2 = 5 \)[/tex], [tex]\( b_2 = 12 \)[/tex], [tex]\( c_2 = -20 \)[/tex].
The distance from [tex]\( P(10, k) \)[/tex] to [tex]\( l_2 \)[/tex] is:
[tex]\[
\text{Distance to } l_2 = \frac{|5(10) + 12(k) - 20|}{\sqrt{5^2 + 12^2}}
\][/tex]
[tex]\[
= \frac{|50 + 12k - 20|}{\sqrt{25 + 144}}
\][/tex]
[tex]\[
= \frac{|30 + 12k|}{13}
\][/tex]
Since [tex]\( P(10, k) \)[/tex] is on the bisector, the distances are equal:
[tex]\[
\frac{|18 - 4k|}{5} = \frac{|30 + 12k|}{13}
\][/tex]
To solve for [tex]\( k \)[/tex], cross-multiply and simplify:
[tex]\[
13|18 - 4k| = 5|30 + 12k|
\][/tex]
Now solve these equations separately for [tex]\( k \)[/tex]:
1. [tex]\( 13(18 - 4k) = 5(30 + 12k) \)[/tex]
2. [tex]\( 13(18 - 4k) = -5(30 + 12k) \)[/tex]
Solution of 1:
[tex]\[
234 - 52k = 150 + 60k
\][/tex]
[tex]\[
234 - 150 = 60k + 52k
\][/tex]
[tex]\[
84 = 112k
\][/tex]
[tex]\[
k = \frac{84}{112} = \frac{3}{4}
\][/tex]
Solution of 2:
[tex]\[
234 - 52k = -150 - 60k
\][/tex]
[tex]\[
234 + 150 = 52k - 60k
\][/tex]
[tex]\[
384 = -8k
\][/tex]
[tex]\[
k = -\frac{384}{8} = -48
\][/tex]
The possible values of [tex]\( k \)[/tex] are [tex]\( \frac{3}{4} \)[/tex] and [tex]\(-48\)[/tex].
### (ii) (a)
The possible values of [tex]\( k \)[/tex] are [tex]\( k = \frac{3}{4} \)[/tex] and [tex]\( k = -48 \)[/tex].
### (ii) (b) If [tex]\( k > 0 \)[/tex], find the distance from [tex]\( P \)[/tex] to [tex]\( l_1 \)[/tex].
Since [tex]\( k = \frac{3}{4} \)[/tex] is the only positive value:
Substitute [tex]\( k = \frac{3}{4} \)[/tex] into the distance formula for [tex]\( l_1 \)[/tex]:
[tex]\[
\text{Distance} = \frac{|18 - 4(\frac{3}{4})|}{5}
\][/tex]
[tex]\[
= \frac{|18 - 3|}{5}
\][/tex]
[tex]\[
= \frac{15}{5}
\][/tex]
[tex]\[
= 3
\][/tex]
So, the distance from [tex]\( P \)[/tex] to [tex]\( l_1 \)[/tex] when [tex]\( k > 0 \)[/tex] is 3.
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