To find the distance between two points in a coordinate plane, we use the distance formula:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
where \((x_1, y_1)\) and \((x_2, y_2)\) are the coordinates of the two points.
Let's apply this formula to each pair of points given in the question.
### (i) A(9, 3) and B(15, 11)
1. Identify the coordinates:
- \(A(9, 3)\) → \(x_1 = 9\), \(y_1 = 3\)
- \(B(15, 11)\) → \(x_2 = 15\), \(y_2 = 11\)
2. Substitute into the distance formula:
\[
d = \sqrt{(15 - 9)^2 + (11 - 3)^2}
\]
\[
= \sqrt{(6)^2 + (8)^2}
\]
\[
= \sqrt{36 + 64}
\]
\[
= \sqrt{100}
\]
\[
= 10
\]
### (ii) A(7, -4) and B(-5, 1)
1. Identify the coordinates:
- \(A(7, -4)\) → \(x_1 = 7\), \(y_1 = -4\)
- \(B(-5, 1)\) → \(x_2 = -5\), \(y_2 = 1\)
2. Substitute into the distance formula:
\[
d = \sqrt{(-5 - 7)^2 + (1 - (-4))^2}
\]
\[
= \sqrt{(-12)^2 + (5)^2}
\]
\[
= \sqrt{144 + 25}
\]
\[
= \sqrt{169}
\]
\[
= 13
\]
### (iii) A(-6, -4) and B(9, -12)
1. Identify the coordinates:
- \(A(-6, -4)\) → \(x_1 = -6\), \(y_1 = -4\)
- \(B(9, -12)\) → \(x_2 = 9\), \(y_2 = -12\)
2. Substitute into the distance formula:
\[
d = \sqrt{(9 - (-6))^2 + (-12 - (-4))^2}
\]
\[
= \sqrt{(15)^2 + (-8)^2}
\]
\[
= \sqrt{225 + 64}
\]
\[
= \sqrt{289}
\]
\[
= 17
\]
### (iv) A(1, -3) and B(4, -6)
1. Identify the coordinates:
- \(A(1, -3)\) → \(x_1 = 1\), \(y_1 = -3\)
- \(B(4, -6)\) → \(x_2 = 4\), \(y_2 = -6\)
2. Substitute into the distance formula:
\[
d = \sqrt{(4 - 1)^2 + (-6 - (-3))^2}
\]
\[
= \sqrt{(3)^2 + (-3)^2}
\]
\[
= \sqrt{9 + 9}
\]
\[
= \sqrt{18}
\]
\[
= 3\sqrt{2}
\]
### (v) P(a+b, a-b) and Q(a-b, a+b)
1. Identify the coordinates:
- \(P(a+b, a-b)\) → \(x_1 = a+b\), \(y_1 = a-b\)
- \(Q(a-b, a+b)\) → \(x_2 = a-b\), \(y_2 = a+b\)
2. Substitute into the distance formula:
\[
d = \sqrt{((a-b) - (a+b))^2 + ((a+b) - (a-b))^2}
\]
\[
= \sqrt{(-2b)^2 + (2a)^2}
\]
\[
= \sqrt{4b^2 + 4a^2}
\]
\[
= 2\sqrt{a^2 + b^2}
\]
### (vi) P(a sin α, a cos α) and Q(a cos α, -a sin α)
1. Identify the coordinates:
- \(P(a \sin \alpha, a \cos \alpha)\) → \(x_1 = a \sin \alpha\), \(y_1 = a \cos \alpha\)
- \(Q(a \cos \alpha, -a \sin \alpha)\) → \(x_2 = a \cos \alpha\), \(y_2 = -a \sin \alpha\)
2. Substitute into the distance formula:
\[
d = \sqrt{(a \cos \alpha - a \sin \alpha)^2 + (-a \sin \alpha - a \cos \alpha)^2}
\]
\[
= \sqrt{(a(\cos \alpha - \sin \alpha))^2 + (-a(\sin \alpha + \cos \alpha))^2}
\]
\[
= a\sqrt{(\cos \alpha - \sin \alpha)^2 + (\sin \alpha + \cos \alpha)^2}
\]
\[
= a\sqrt{(\cos^2 \alpha - 2\sin \alpha \cos \alpha + \sin^2 \alpha) + (\sin^2 \alpha + 2\sin \alpha \cos \alpha + \cos^2 \alpha)}
\]
\[
= a\sqrt{2(\cos^2 \alpha + \sin^2 \alpha)}
\]
\[
= a\sqrt{2}
\]
### Summary of Distances:
1. Distance between A(9, 3) and B(15, 11) = 10 units
2. Distance between A(7, -4) and B(-5, 1) = 13 units
3. Distance between A(-6, -4) and B(9, -12) = 17 units
4. Distance between A(1, -3) and B(4, -6) = \(3\sqrt{2}\) units
5. Distance between P(a+b, a-b) and Q(a-b, a+b) = \(2\sqrt{a^2 + b^2}\) units
6. Distance between P(a sin α, a cos α) and Q(a cos α, -a sin α) = \(a\sqrt{2}\) units
