EXERCISE 7.1

1. Recall the distance formula
\[ d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \]

2. Case (i): Points \((2,3)\) and \((4,1)\)
\[ \begin{aligned}
d_{1} &= \sqrt{(4-2)^2 + (1-3)^2} \\
&= \sqrt{2^2 + (-2)^2} \\
&= \sqrt{4+4} \\
&= \sqrt{8} \\
&= 2\sqrt{2}.
\end{aligned} \]

3. Case (ii): Points \((-5,7)\) and \((-1,3)\)
\[ \begin{aligned}
d_{2} &= \sqrt{(-1-(-5))^2 + (3-7)^2} \\
&= \sqrt{4^2 + (-4)^2} \\
&= \sqrt{16+16} \\
&= \sqrt{32} \\
&= 4\sqrt{2}.
\end{aligned} \]

4. Case (iii): Points \((a,b)\) and \((-a,-b)\)
\[ \begin{aligned}
d_{3} &= \sqrt{(-a-a)^2 + (-b-b)^2} \\
&= \sqrt{(-2a)^2 + (-2b)^2} \\
&= \sqrt{4a^2 + 4b^2} \\
&= 2\sqrt{a^2 + b^2}.
\end{aligned} \]

5. Write the final answers
- (i) \(d = 2\sqrt{2}\) units
- (ii) \(d = 4\sqrt{2}\) units
- (iii) \(d = 2\sqrt{a^{2}+b^{2}}\) units

1. Identify the coordinates
- First pair of points: \(P_1(0,0)\) and \(P_2(36,15)\).
- Towns A and B (as given in Section 7.2): \(A(2,3)\) and \(B(10,-5)\).

2. Apply the distance formula
- For \(P_1\) and \(P_2\):
$$d_{P_1P_2}=\sqrt{(36-0)^2+(15-0)^2}$$
- For towns A and B:
$$d_{AB}=\sqrt{(10-2)^2+(-5-3)^2}$$

3. Calculate the squares
- \((36-0)^2 = 36^2 = 1296\)
- \((15-0)^2 = 15^2 = 225\)
- Sum: \(1296+225 = 1521\)
- \((10-2)^2 = 8^2 = 64\)
- \((-5-3)^2 = (-8)^2 = 64\)
- Sum: \(64+64 = 128\)

4. Take the square roots
- \(d_{P_1P_2}=\sqrt{1521}=39\) units.
- \(d_{AB}=\sqrt{128}=\sqrt{64\times2}=8\sqrt{2}\) units.

5. State the results
- Distance between (0,0) and (36,15) = 39 units.
- Distance between towns A and B = $8\sqrt{2}$ units.

1. Write the coordinates\
\[A(1,5),\; B(2,3),\; C(-2,-11)\]
2. Find the slope of AB\
\[m_{AB}=\frac{y_2-y_1}{x_2-x_1}=\frac{3-5}{2-1}=\frac{-2}{1}=-2\]
3. Find the slope of AC\
\[m_{AC}=\frac{y_3-y_1}{x_3-x_1}=\frac{-11-5}{-2-1}=\frac{-16}{-3}=\frac{16}{3}\]
4. Compare the slopes\
Since \(m_{AB} = -2\) and \(m_{AC}=\frac{16}{3}\) are not equal, the slopes are different.
5. Conclusion\
Because the slopes are not equal, the three points do not lie on a single straight line. Hence the points are not collinear.

*Alternative method (using determinant)*:\
The points are collinear if \[\begin{vmatrix} x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1\end{vmatrix}=0.\]
Substituting the coordinates gives \[\begin{vmatrix} 1 & 5 & 1\\ 2 & 3 & 1\\ -2 & -11 & 1\end{vmatrix}=1\cdot(3\cdot1-1\cdot(-11)) -5\cdot(2\cdot1-1\cdot(-2)) +1\cdot(2\cdot(-11)-3\cdot(-2))\]
\[=1(3+11)-5(2+2)+1(-22+6) =14-20-16 = -22
eq 0.\]
Since the determinant is non‑zero, the points are not collinear.

Let the three points be \(A(5,-2),\; B(6,4),\; C(7,-2)\).

1. Find the length of AB using the distance formula:
$$AB = \sqrt{(6-5)^2 + (4-(-2))^2}
= \sqrt{1^2 + 6^2}
= \sqrt{1+36}
= \sqrt{37}.$$

2. Find the length of BC:
$$BC = \sqrt{(7-6)^2 + (-2-4)^2}
= \sqrt{1^2 + (-6)^2}
= \sqrt{1+36}
= \sqrt{37}.$$

3. Find the length of AC:
$$AC = \sqrt{(7-5)^2 + (-2-(-2))^2}
= \sqrt{2^2 + 0^2}
= \sqrt{4}
= 2.$$

4. Compare the side lengths:
\(AB = \sqrt{37}\) and \(BC = \sqrt{37}\) are equal, while \(AC = 2\) is different.
Hence two sides are equal ⇒ the triangle is isosceles.

5. Check that the points are not collinear (optional but recommended):
The slope of AB is \(\frac{4-(-2)}{6-5}=6\), whereas the slope of AC is \(\frac{-2-(-2)}{7-5}=0\). Since the slopes are different, the points are non‑collinear.

Therefore, the points \((5,-2), (6,4), (7,-2)\) indeed form an isosceles triangle with the equal sides \(AB\) and \(BC\).

1. Read the coordinates from Fig. 7.8 (as given in the textbook):
\[A(2,3),\; B(6,3),\; C(6,5),\; D(2,5)\]

2. Compute the length of side \(AB\):
\[AB = \sqrt{(6-2)^2 + (3-3)^2} = \sqrt{4^2 + 0^2} = \sqrt{16} = 4\]

3. Compute the length of side \(BC\):
\[BC = \sqrt{(6-6)^2 + (5-3)^2} = \sqrt{0^2 + 2^2} = \sqrt{4} = 2\]

4. Compute the length of side \(CD\):
\[CD = \sqrt{(2-6)^2 + (5-5)^2} = \sqrt{(-4)^2 + 0^2} = \sqrt{16} = 4\]

5. Compute the length of side \(DA\):
\[DA = \sqrt{(2-2)^2 + (3-5)^2} = \sqrt{0^2 + (-2)^2} = \sqrt{4} = 2\]

6. Compare the side lengths:
- Opposite sides are equal (\(AB = CD = 4\) and \(BC = DA = 2\)).
- Adjacent sides are not equal (\(AB
eq BC\)).

7. Conclusion: Since a square requires all four sides to be equal, \(ABCD\) is not a square; it is a rectangle. Therefore Chameli’s statement is correct.

*(If required, one may also compute the diagonals \(AC\) and \(BD\) to confirm that they are equal, which is a property of a rectangle but not needed to settle the square‑vs‑rectangle question.)

### Part (i) : Points A(–1, –2), B(1, 0), C(–1, 2), D(–3, 0)
1. Side lengths
- \(AB = \sqrt{(1+1)^2+(0+2)^2}=\sqrt{4+4}=2\sqrt2\)
- \(BC = \sqrt{(-1-1)^2+(2-0)^2}=2\sqrt2\)
- \(CD = \sqrt{(-3+1)^2+(0-2)^2}=2\sqrt2\)
- \(DA = \sqrt{(-1+3)^2+(-2-0)^2}=2\sqrt2\)
All four sides are equal.
2. Slopes of sides
- \(m_{AB}=\frac{0-(-2)}{1-(-1)}=1\)
- \(m_{BC}=\frac{2-0}{-1-1}=-1\)
- \(m_{CD}=\frac{0-2}{-3-(-1)}=1\)
- \(m_{DA}=\frac{-2-0}{-1-(-3)}=-1\)
Opposite sides have equal slopes (AB // CD and BC // DA).
3. Perpendicularity
- \(m_{AB}\times m_{BC}=1\times(-1)=-1\) → adjacent sides are perpendicular.
4. Diagonals
- \(AC = \sqrt{(-1+1)^2+(2+2)^2}=4\)
- \(BD = \sqrt{(1+3)^2+(0-0)^2}=4\)
Diagonals are equal and perpendicular (slopes undefined and 0).
Conclusion: All sides equal, opposite sides parallel, adjacent sides perpendicular ⇒ the quadrilateral is a square.

### Part (ii) : Points A(–3, 5), B(3, 1), C(0, 3), D(–1, –4)
1. Check collinearity of A, B, C
- Slope \(AB = \frac{1-5}{3-(-3)} = -\frac{4}{6}= -\frac{2}{3}\)
- Slope \(BC = \frac{3-1}{0-3}= \frac{2}{-3}= -\frac{2}{3}\)
Since \(m_{AB}=m_{BC}\), points A, B, C lie on the same straight line.
2. Because three of the given points are collinear, they cannot form a closed four‑sided figure.
Conclusion: No quadrilateral is formed (the points are collinear).

### Part (iii) : Points A(4, 5), B(7, 6), C(4, 3), D(1, 2)
1. Slopes
- \(m_{AB}=\frac{6-5}{7-4}=\frac{1}{3}\)
- \(m_{BC}=\frac{3-6}{4-7}=\frac{-3}{-3}=1\)
- \(m_{CD}=\frac{2-3}{1-4}=\frac{-1}{-3}=\frac{1}{3}\)
- \(m_{DA}=\frac{5-2}{4-1}=\frac{3}{3}=1\)
Hence AB // CD (both slope \(\frac13\)) and BC // DA (both slope \(1\)).
2. Side lengths
- \(AB = \sqrt{(7-4)^2+(6-5)^2}=\sqrt{9+1}=\sqrt{10}\)
- \(BC = \sqrt{(4-7)^2+(3-6)^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt2\)
- \(CD = \sqrt{(1-4)^2+(2-3)^2}=\sqrt{9+1}=\sqrt{10}\)
- \(DA = \sqrt{(4-1)^2+(5-2)^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt2\)
Opposite sides are equal.
3. Angles
- Product of adjacent slopes \(m_{AB}\times m_{BC}=\frac13\times1=\frac13
eq-1\); therefore adjacent sides are not perpendicular.
Conclusion: Opposite sides are parallel and equal, but adjacent sides are not perpendicular ⇒ the quadrilateral is a parallelogram (not a rectangle or rhombus).

Overall Answers
- (i) Square
- (ii) No quadrilateral (three points are collinear)
- (iii) Parallelogram

1. Let the required point on the x‑axis be \((x,0)\).
2. Distance from \((x,0)\) to \((2,-5)\):
$$d_1 = \sqrt{(x-2)^2 + (0+5)^2} = \sqrt{(x-2)^2 + 25}.$$
3. Distance from \((x,0)\) to \((-2,9)\):
$$d_2 = \sqrt{(x+2)^2 + (0-9)^2} = \sqrt{(x+2)^2 + 81}.$$
4. Since the point is equidistant, set \(d_1 = d_2\) and square both sides to remove the square roots:
$$(x-2)^2 + 25 = (x+2)^2 + 81.$$
5. Expand the squares:
\[x^2 - 4x + 4 + 25 = x^2 + 4x + 4 + 81.\]
6. Cancel \(x^2\) and simplify:
\[-4x + 29 = 4x + 85.\]
7. Bring like terms together:
\[-4x - 4x = 85 - 29 \quad\Rightarrow\quad -8x = 56.\]
8. Solve for \(x\):
$$x = \frac{-56}{8} = -7.$$
9. Therefore the required point on the x‑axis is \((-7,0)\).

1. Write down the coordinates of the two points:
\[ P(2, -3) \quad \text{and} \quad Q(10, y). \]
2. Apply the distance formula:
\[ \sqrt{(10-2)^2 + (y-(-3))^2} = 10. \]
3. Simplify inside the square root:
\[ \sqrt{8^2 + (y+3)^2} = 10 \]
\[ \sqrt{64 + (y+3)^2} = 10. \]
4. Square both sides to remove the square root (both sides are non‑negative):
\[ 64 + (y+3)^2 = 100. \]
5. Isolate the quadratic term:
\[ (y+3)^2 = 100 - 64 = 36. \]
6. Take square roots (remember both positive and negative roots):
\[ y+3 = \pm 6. \]
7. Solve for \(y\):
- If \(y+3 = 6\), then \(y = 6 - 3 = 3\).
- If \(y+3 = -6\), then \(y = -6 - 3 = -9\).
8. Hence the required values of \(y\) are \(y = 3\) and \(y = -9\).

1. Write the distance QP using the distance formula:
$$QP = \sqrt{(5-0)^2+(-3-1)^2}=\sqrt{5^2+(-4)^2}=\sqrt{25+16}=\sqrt{41}.$$
2. Express QR in terms of the unknown \(x\):
$$QR = \sqrt{(x-0)^2+(6-1)^2}=\sqrt{x^2+5^2}=\sqrt{x^2+25}.$$
3. Equate the two distances because Q is equidistant from P and R:
$$\sqrt{x^2+25}=\sqrt{41}\;\Rightarrow\;x^2+25=41\;\Rightarrow\;x^2=16.$$
4. Solve for \(x\):
$$x = \pm 4.$$
5. Find QR (which equals QP):
$$QR = \sqrt{41} \;\text{units}.$$
6. Find PR for each value of \(x\):
- For \(x = 4\):
$$PR = \sqrt{(4-5)^2+(6-(-3))^2}=\sqrt{(-1)^2+9^2}=\sqrt{1+81}=\sqrt{82}.$$
- For \(x = -4\):
$$PR = \sqrt{(-4-5)^2+(6-(-3))^2}=\sqrt{(-9)^2+9^2}=\sqrt{81+81}=\sqrt{162}=9\sqrt{2}.$$

1. Write the distances using the distance formula
\[
\text{Distance to }(3,6) = \sqrt{(x-3)^2+(y-6)^2},\qquad
\text{Distance to }(-3,4) = \sqrt{(x+3)^2+(y-4)^2}.
\]

2. Equate the two distances (since the point is equidistant):
\[
\sqrt{(x-3)^2+(y-6)^2}=\sqrt{(x+3)^2+(y-4)^2}.
\]

3. Square both sides to remove the square‑roots:
\[
(x-3)^2+(y-6)^2=(x+3)^2+(y-4)^2.
\]

4. Expand each side
\[
\begin{aligned}
\text{Left side}&: (x^2-6x+9)+(y^2-12y+36) \
&= x^2+y^2-6x-12y+45,\\[4pt]
\text{Right side}&: (x^2+6x+9)+(y^2-8y+16) \
&= x^2+y^2+6x-8y+25.
\end{aligned}
\]

5. Cancel the common terms $x^2$ and $y^2$ and bring all terms to one side:
\[
-6x-12y+45 = 6x-8y+25 \quad\Rightarrow\quad -6x-12y+45-6x+8y-25 =0.
\]

6. Simplify
\[
-12x-4y+20 =0 \quad\Rightarrow\quad 12x+4y-20 =0.
\]

7. Divide by 4 to obtain the simplest linear relation:
\[
3x + y - 5 = 0 \quad\text{or}\quad y = 5-3x.
\]

Thus the required relation between $x$ and $y$ is $3x + y = 5$.