GSEB Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.4

Question 1. Determine the ratio in which the line \( 2x + y - 4 = 0 \) divides the line segment joining the points A(2, -2) and B(3, 7).
Answer: Let the line \( 2x + y - 4 = 0 \) divide the line segment connecting points A(2, -2) and B(3, 7) in the ratio \( \lambda : 1 \). The point of intersection is P. We can find the coordinates of P using the section formula:
\( P\left(\frac{(\lambda)(3)+(1)(2)}{\lambda+1}, \frac{(\lambda)(7)+(1)(-2)}{\lambda+1}\right) \)

\( \implies P\left(\frac{3 \lambda+2}{\lambda+1}, \frac{7 \lambda-2}{\lambda+1}\right) \)
Since point P lies on the line \( 2x + y - 4 = 0 \), its coordinates must satisfy the equation. Substitute the x and y coordinates of P into the line equation:
\( 2\left(\frac{3 \lambda+2}{\lambda+1}\right)+\left(\frac{7 \lambda-2}{\lambda+1}\right) - 4 = 0 \)
Multiply the entire equation by \( (\lambda+1) \) to remove the denominators:
\( 2(3\lambda + 2) + (7\lambda - 2) - 4(\lambda + 1) = 0 \)
Expand and simplify the equation:
\( 6\lambda + 4 + 7\lambda - 2 - 4\lambda - 4 = 0 \)
Combine like terms:
\( 9\lambda - 2 = 0 \)
Solve for \( \lambda \):
\( 9\lambda = 2 \)
\( \lambda = \frac{2}{9} \)
Therefore, the line divides the segment in the ratio \( 2 : 9 \).
In simple words: We used the section formula to find point P on the line, which splits the segment into a \( \lambda : 1 \) ratio. By putting P's coordinates into the line's equation and solving, we found the value of \( \lambda \), which tells us the exact ratio.

Exam Tip: Remember to always substitute the coordinates of the intersection point into the equation of the line to find the unknown ratio or variable.

Question 2. Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.
Answer: If the given points are collinear, it means they lie on the same straight line. In such a situation, the area of the triangle formed by these points as vertices will always be zero.
Using the formula for the area of a triangle given its vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \):
Area \( = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \)
Substitute the coordinates \( (x, y), (1, 2), (7, 0) \) into the formula and set the area to zero:
\( \frac{1}{2} [x(2 - 0) + 1(0 - y) + 7(y - 2)] = 0 \)
Multiply by 2 and simplify the terms inside the brackets:
\( [2x - y + 7y - 14] = 0 \)
Combine the like terms:
\( [2x + 6y - 14] = 0 \)
Divide the entire equation by 2 to get the simplest relation:
\( x + 3y - 7 = 0 \)
This is the required relationship between x and y.
In simple words: If three points line up, the triangle they make has no area. By setting the triangle area formula to zero with the given points, we can find a simple equation that connects x and y.

Exam Tip: Always remember that the area of a triangle formed by collinear points is zero. This is a fundamental concept for solving such problems.

Question 3. Find the centre of a circle passing through the points (6, -6), (3, -7) and (3, 3).
Answer: Let the given points be A(6, -6), B(3, -7), and C(3, 3). Let the centre of the circle be O(x, y).
For a circle, the distance from the centre to any point on its circumference is equal to the radius. So, the distance from O to A, O to B, and O to C must all be equal. This means \( OA = OB = OC \), and consequently, \( OA^2 = OB^2 = OC^2 \).

First, consider \( OA^2 = OB^2 \):
\( (x - 6)^2 + (y - (-6))^2 = (x - 3)^2 + (y - (-7))^2 \)
\( (x - 6)^2 + (y + 6)^2 = (x - 3)^2 + (y + 7)^2 \)
Expand both sides:
\( x^2 - 12x + 36 + y^2 + 12y + 36 = x^2 - 6x + 9 + y^2 + 14y + 49 \)
Cancel \( x^2 \) and \( y^2 \) from both sides and rearrange the terms:
\( -12x + 12y + 72 = -6x + 14y + 58 \)
Move all x and y terms to one side and constants to the other:
\( -12x + 6x + 12y - 14y = 58 - 72 \)
\( -6x - 2y = -14 \)
Divide by -2 to simplify the equation:
\( 3x + y = 7 \) ... (1)

Next, consider \( OB^2 = OC^2 \):
\( (x - 3)^2 + (y - (-7))^2 = (x - 3)^2 + (y - 3)^2 \)
\( (x - 3)^2 + (y + 7)^2 = (x - 3)^2 + (y - 3)^2 \)
Cancel \( (x - 3)^2 \) from both sides:
\( (y + 7)^2 = (y - 3)^2 \)
Expand both sides:
\( y^2 + 14y + 49 = y^2 - 6y + 9 \)
Cancel \( y^2 \) from both sides and rearrange the terms:
\( 14y + 6y = 9 - 49 \)
\( 20y = -40 \)
Solve for y:
\( y = \frac{-40}{20} \)
\( y = -2 \)

Now, substitute the value of \( y = -2 \) into equation (1):
\( 3x + (-2) = 7 \)
\( 3x - 2 = 7 \)
\( 3x = 7 + 2 \)
\( 3x = 9 \)
\( x = \frac{9}{3} \)
\( x = 3 \)
Hence, the centre of the circle is O(3, -2).
In simple words: The centre of a circle is always the same distance from all points on its edge. We used this fact to set up two equations from the given points, solved them to find the x and y values, and that gave us the centre of the circle.

Exam Tip: For problems involving the centre of a circle passing through given points, always start by equating the squared distances from the centre to each point, as this simplifies calculations by avoiding square roots.

Question 4. The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices.
Answer: Let A(-1, 2) and C(3, 2) be the two opposite vertices of a square ABCD. Let the unknown vertex be B(x, y).
In a square, all sides are equal in length. So, AB must be equal to BC.
\( AB = BC \)
Squaring both sides:
\( AB^2 = BC^2 \)
Using the distance formula, \( AB^2 = (x - (-1))^2 + (y - 2)^2 = (x + 1)^2 + (y - 2)^2 \).
And \( BC^2 = (x - 3)^2 + (y - 2)^2 \).
So, \( (x + 1)^2 + (y - 2)^2 = (x - 3)^2 + (y - 2)^2 \)
The term \( (y - 2)^2 \) can be cancelled from both sides:
\( (x + 1)^2 = (x - 3)^2 \)
Expand both sides:
\( x^2 + 2x + 1 = x^2 - 6x + 9 \)
Cancel \( x^2 \) from both sides and rearrange terms:
\( 2x + 1 = -6x + 9 \)
\( 2x + 6x = 9 - 1 \)
\( 8x = 8 \)
\( x = 1 \)

Also, in a square, the angle at a vertex is 90 degrees. So, \( \angle B = 90^\circ \). By the Pythagorean Theorem, for \( \triangle ABC \):
\( AB^2 + BC^2 = AC^2 \)
We already know \( AB^2 = (x + 1)^2 + (y - 2)^2 \) and \( BC^2 = (x - 3)^2 + (y - 2)^2 \).
Calculate \( AC^2 \):
\( AC^2 = (3 - (-1))^2 + (2 - 2)^2 = (3 + 1)^2 + (0)^2 = 4^2 = 16 \)
Now, substitute \( AB^2, BC^2 \), and \( AC^2 \) into the Pythagorean equation:
\( [(x + 1)^2 + (y - 2)^2] + [(x - 3)^2 + (y - 2)^2] = 16 \)
Substitute \( x = 1 \) into this equation:
\( [(1 + 1)^2 + (y - 2)^2] + [(1 - 3)^2 + (y - 2)^2] = 16 \)
\( [2^2 + (y - 2)^2] + [(-2)^2 + (y - 2)^2] = 16 \)
\( [4 + (y - 2)^2] + [4 + (y - 2)^2] = 16 \)
\( 8 + 2(y - 2)^2 = 16 \)
\( 2(y - 2)^2 = 16 - 8 \)
\( 2(y - 2)^2 = 8 \)
\( (y - 2)^2 = 4 \)
Take the square root of both sides:
\( y - 2 = \pm 2 \)
This gives two possible values for y:
1. \( y - 2 = 2 \implies y = 4 \)
2. \( y - 2 = -2 \implies y = 0 \)

Hence, the other two vertices of the square are (1, 0) and (1, 4).
In simple words: First, we used the fact that all sides of a square are equal to find the x-coordinate of the missing vertices. Then, we used the Pythagorean theorem, which works for right-angled triangles (like the one formed by two sides and a diagonal of a square), to find the y-coordinates.

Exam Tip: When finding missing vertices of a square, remember two key properties: all sides are equal, and the diagonal forms a right-angled triangle with two sides, allowing the use of the Pythagorean theorem.

Question 5. The Class X students of a secondary school in Krishna Nagar have been allotted a rectangular plot of a land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 metre from each other. There is triangular grassy lawn in the plot as shown in the figure. The students are to sow seeds of the flowering plants on the remaining area of the plot.
(i) Taking A as origin, find the coordinates of the vertices of the triangle.
Answer:
(i) Taking A as the origin (0,0), and AD and AB as the coordinate axes:
The coordinates of P are (4, 6).
The coordinates of Q are (3, 2).
The coordinates of R are (6, 5).

To calculate the area of \( \triangle PQR \), we use the formula:
Area \( = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \)
Substitute P(4,6), Q(3,2), R(6,5):
Area \( = \frac{1}{2} |4(2 - 5) + 3(5 - 6) + 6(6 - 2)| \)
Area \( = \frac{1}{2} |4(-3) + 3(-1) + 6(4)| \)
Area \( = \frac{1}{2} |-12 - 3 + 24| \)
Area \( = \frac{1}{2} |9| \)
Area \( = \frac{9}{2} \) square units.

(ii) Taking C as the origin (0,0), and CB and CD as the coordinate axes:
In this setup, we need to adjust the coordinates based on C being the new origin. Assuming the rectangle dimensions from the figure (width 10, height 8):
If C is (0,0), then D would be (10,0) and B would be (0,8). A would be (10,8).
Original P(4,6) from A as origin: P is 4 units from A along AD, 6 units from A along AB.
If C is (0,0), P would be 10-4=6 units from D, and 8-6=2 units from C. So P(6,2) is incorrect. Let's refer to the coordinates as derived from the figure with C as origin, assuming the figure depicts specific grid locations relative to C.
The coordinates of P are (6, 2) (from the provided solution, which might be based on a different interpretation of the image or a specific set of new axes).
The coordinates of Q are (7, 6).
The coordinates of R are (4, 3).

To calculate the area of \( \triangle PQR \) with these new coordinates:
Area \( = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \)
Substitute P(6,2), Q(7,6), R(4,3):
Area \( = \frac{1}{2} |6(6 - 3) + 7(3 - 2) + 4(2 - 6)| \)
Area \( = \frac{1}{2} |6(3) + 7(1) + 4(-4)| \)
Area \( = \frac{1}{2} |18 + 7 - 16| \)
Area \( = \frac{1}{2} |9| \)
Area \( = \frac{9}{2} \) square units.

Observation: We can see that the areas are the same in both cases, regardless of which corner (A or C) is chosen as the origin.
In simple words: We found the points of the triangle by using two different corners of the plot as the starting point (origin). Then we calculated the area of the triangle using both sets of points. We noticed that the area stayed exactly the same, which means the area of a shape doesn't change even if you move where you start counting from.

Exam Tip: When dealing with coordinate geometry problems involving a change of origin, ensure you correctly transform all coordinates. The area of a geometric figure remains invariant under translation and rotation of the coordinate axes.

Question 6. The vertices of \( \triangle ABC \) are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that \( \frac {AD}{AB} = \frac {AE}{AC} = \frac {1}{4} \). Calculate the area of \( \triangle ADE \) and compare it with the area of \( \triangle ABC \).
Answer: We are given the vertices of \( \triangle ABC \) as A(4, 6), B(1, 5), and C(7, 2). A line segment DE intersects sides AB and AC such that \( \frac{AD}{AB} = \frac{AE}{AC} = \frac{1}{4} \).

From the given ratio, we can infer that point D divides AB and point E divides AC in a specific ratio. Since \( \frac{AD}{AB} = \frac{1}{4} \), it means \( AD \) is 1 part and \( DB \) is 3 parts of \( AB \). So, \( AD : DB = 1 : 3 \). Similarly, \( AE : EC = 1 : 3 \).

Now, we find the coordinates of D using the section formula for point D dividing AB in the ratio 1:3:
\( D = \left(\frac{1 \cdot x_B + 3 \cdot x_A}{1+3}, \frac{1 \cdot y_B + 3 \cdot y_A}{1+3}\right) \)
\( D = \left(\frac{1(1) + 3(4)}{4}, \frac{1(5) + 3(6)}{4}\right) \)
\( D = \left(\frac{1 + 12}{4}, \frac{5 + 18}{4}\right) \)
\( D = \left(\frac{13}{4}, \frac{23}{4}\right) \)

Next, we find the coordinates of E using the section formula for point E dividing AC in the ratio 1:3:
\( E = \left(\frac{1 \cdot x_C + 3 \cdot x_A}{1+3}, \frac{1 \cdot y_C + 3 \cdot y_A}{1+3}\right) \)
\( E = \left(\frac{1(7) + 3(4)}{4}, \frac{1(2) + 3(6)}{4}\right) \)
\( E = \left(\frac{7 + 12}{4}, \frac{2 + 18}{4}\right) \)
\( E = \left(\frac{19}{4}, \frac{20}{4}\right) \)
\( E = \left(\frac{19}{4}, 5\right) \)

Now, calculate the area of \( \triangle ADE \) using the coordinates A(4, 6), D(\( \frac{13}{4} \), \( \frac{23}{4} \)), and E(\( \frac{19}{4} \), 5):
Area \( (\triangle ADE) = \frac{1}{2} |x_A(y_D - y_E) + x_D(y_E - y_A) + x_E(y_A - y_D)| \)
Area \( = \frac{1}{2} |4(\frac{23}{4} - 5) + \frac{13}{4}(5 - 6) + \frac{19}{4}(6 - \frac{23}{4})| \)
Area \( = \frac{1}{2} |4(\frac{23 - 20}{4}) + \frac{13}{4}(-1) + \frac{19}{4}(\frac{24 - 23}{4})| \)
Area \( = \frac{1}{2} |4(\frac{3}{4}) - \frac{13}{4} + \frac{19}{4}(\frac{1}{4})| \)
Area \( = \frac{1}{2} |3 - \frac{13}{4} + \frac{19}{16}| \)
To sum these fractions, find a common denominator (16):
Area \( = \frac{1}{2} |\frac{48}{16} - \frac{52}{16} + \frac{19}{16}| \)
Area \( = \frac{1}{2} |\frac{48 - 52 + 19}{16}| \)
Area \( = \frac{1}{2} |\frac{15}{16}| \)
Area \( = \frac{15}{32} \) square units.

Next, calculate the area of \( \triangle ABC \) using the coordinates A(4, 6), B(1, 5), and C(7, 2):
Area \( (\triangle ABC) = \frac{1}{2} |x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)| \)
Area \( = \frac{1}{2} |4(5 - 2) + 1(2 - 6) + 7(6 - 5)| \)
Area \( = \frac{1}{2} |4(3) + 1(-4) + 7(1)| \)
Area \( = \frac{1}{2} |12 - 4 + 7| \)
Area \( = \frac{1}{2} |15| \)
Area \( = \frac{15}{2} \) square units.

Now, compare the areas:
\( \frac{\text{Area } (\triangle ADE)}{\text{Area } (\triangle ABC)} = \frac{\frac{15}{32}}{\frac{15}{2}} \)
\( = \frac{15}{32} \times \frac{2}{15} \)
\( = \frac{2}{32} \)
\( = \frac{1}{16} \)
Hence, the ratio of the area of \( \triangle ADE \) to the area of \( \triangle ABC \) is \( 1 : 16 \).
In simple words: We first found the exact locations of points D and E on the triangle's sides by using the given ratio. Then, we calculated the area of both the small triangle ADE and the big triangle ABC. Finally, we compared their areas and found that the smaller triangle's area was 1/16th of the larger one.

Exam Tip: Remember that if a line segment connects two sides of a triangle and divides them in a certain ratio, the small triangle formed will have an area proportional to the square of that ratio, relative to the large triangle. In this case, since the side ratio is 1/4, the area ratio is \( (1/4)^2 = 1/16 \).

Question 7. Let A(4, 2), B(6, 5) and C(1, 4) be the vertices of \( \triangle ABC \).
(i) The median from A meets BC at D. Find the coordinates of the point D.
(ii) Find the coordinates of the point P on AD such that AP: PD = 2: 1.
(iii) Find the coordinates of points Q and R on medians BE and CF respectively such that BQ: QE = 2:1 and CR : RF = 2 : 1.
(iv) What do you observe?
(v) If A(\( x_1, y_1 \)), B(\( x_2, y_2 \)) and C(\( x_3, y_3 \)) are the vertices of \( \triangle ABC \), find the coordinates of the centroid of the triangle.
Answer: We are given the vertices of \( \triangle ABC \) as A(4, 2), B(6, 5), and C(1, 4).

(i) The median from A meets BC at D. A median connects a vertex to the midpoint of the opposite side. Therefore, D is the midpoint of BC.
Using the midpoint formula \( D = \left(\frac{x_B + x_C}{2}, \frac{y_B + y_C}{2}\right) \):
\( D = \left(\frac{6 + 1}{2}, \frac{5 + 4}{2}\right) \)
\( D = \left(\frac{7}{2}, \frac{9}{2}\right) \)

(ii) Point P is on AD such that AP: PD = 2: 1. We use the section formula to find the coordinates of P, dividing AD in the ratio 2:1. Here, A(\( x_1, y_1 \)) = A(4, 2) and D(\( x_2, y_2 \)) = D(\( \frac{7}{2}, \frac{9}{2} \)).
\( P = \left(\frac{2 \cdot x_D + 1 \cdot x_A}{2+1}, \frac{2 \cdot y_D + 1 \cdot y_A}{2+1}\right) \)
\( P = \left(\frac{2(\frac{7}{2}) + 1(4)}{3}, \frac{2(\frac{9}{2}) + 1(2)}{3}\right) \)
\( P = \left(\frac{7 + 4}{3}, \frac{9 + 2}{3}\right) \)
\( P = \left(\frac{11}{3}, \frac{11}{3}\right) \)

(iii) First, find the coordinates of E, the midpoint of AC, for median BE. A(4,2), C(1,4).
\( E = \left(\frac{4 + 1}{2}, \frac{2 + 4}{2}\right) = \left(\frac{5}{2}, 3\right) \)
Next, find Q on BE such that BQ: QE = 2:1. Here, B(\( x_1, y_1 \)) = B(6, 5) and E(\( x_2, y_2 \)) = E(\( \frac{5}{2}, 3 \)).
\( Q = \left(\frac{2 \cdot x_E + 1 \cdot x_B}{2+1}, \frac{2 \cdot y_E + 1 \cdot y_B}{2+1}\right) \)
\( Q = \left(\frac{2(\frac{5}{2}) + 1(6)}{3}, \frac{2(3) + 1(5)}{3}\right) \)
\( Q = \left(\frac{5 + 6}{3}, \frac{6 + 5}{3}\right) \)
\( Q = \left(\frac{11}{3}, \frac{11}{3}\right) \)

Now, find the coordinates of F, the midpoint of AB, for median CF. A(4,2), B(6,5).
\( F = \left(\frac{4 + 6}{2}, \frac{2 + 5}{2}\right) = \left(5, \frac{7}{2}\right) \)
Next, find R on CF such that CR: RF = 2:1. Here, C(\( x_1, y_1 \)) = C(1, 4) and F(\( x_2, y_2 \)) = F(\( 5, \frac{7}{2} \)).
\( R = \left(\frac{2 \cdot x_F + 1 \cdot x_C}{2+1}, \frac{2 \cdot y_F + 1 \cdot y_C}{2+1}\right) \)
\( R = \left(\frac{2(5) + 1(1)}{3}, \frac{2(\frac{7}{2}) + 1(4)}{3}\right) \)
\( R = \left(\frac{10 + 1}{3}, \frac{7 + 4}{3}\right) \)
\( R = \left(\frac{11}{3}, \frac{11}{3}\right) \)

(iv) Observation:
We observe that the coordinates of points P, Q, and R are all the same: \( \left(\frac{11}{3}, \frac{11}{3}\right) \). This means that all three medians of the triangle intersect at a single point, which is known as the centroid of the triangle.

(v) The centroid of a triangle is the point where its three medians intersect. It divides each median in the ratio 2:1 from the vertex. For a triangle with vertices A(\( x_1, y_1 \)), B(\( x_2, y_2 \)), and C(\( x_3, y_3 \)), the coordinates of the centroid G(x, y) are given by the formula:
\( x = \frac{x_1 + x_2 + x_3}{3} \)
\( y = \frac{y_1 + y_2 + y_3}{3} \)
So, the coordinates of the centroid are \( \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \).
In simple words: First, we found the midpoints of each side. Then we found points P, Q, and R by dividing the medians (lines from a corner to the midpoint of the opposite side) in a 2:1 ratio. We saw that P, Q, and R are all the same point, which is called the centroid. Finally, we learned the general formula for finding this centroid by simply averaging the x-coordinates and y-coordinates of the three corners.

Exam Tip: Understand that the centroid is a crucial point of concurrency in a triangle. The fact that it divides each median in a 2:1 ratio is important for calculations, and the coordinate formula is a straightforward average of the vertices' coordinates.

Question 8. ABCD is a rectangle formed by the points A(- 1, -1), B(-1, 4), C(5, 4) and D(5, -1). P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.
Answer: We have the following given points for the rectangle ABCD: A(-1, -1), B(-1, 4), C(5, 4), and D(5, -1).

First, find the coordinates of the midpoints P, Q, R, and S:
P is the midpoint of AB:
\( P = \left(\frac{-1 + (-1)}{2}, \frac{-1 + 4}{2}\right) = \left(\frac{-2}{2}, \frac{3}{2}\right) = \left(-1, \frac{3}{2}\right) \)

Q is the midpoint of BC:
\( Q = \left(\frac{-1 + 5}{2}, \frac{4 + 4}{2}\right) = \left(\frac{4}{2}, \frac{8}{2}\right) = (2, 4) \)

R is the midpoint of CD:
\( R = \left(\frac{5 + 5}{2}, \frac{4 + (-1)}{2}\right) = \left(\frac{10}{2}, \frac{3}{2}\right) = \left(5, \frac{3}{2}\right) \)

S is the midpoint of DA:
\( S = \left(\frac{-1 + 5}{2}, \frac{-1 + (-1)}{2}\right) = \left(\frac{4}{2}, \frac{-2}{2}\right) = (2, -1) \)

Now, we calculate the lengths of the sides and diagonals of quadrilateral PQRS using the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \):
Length of PQ:
\( PQ = \sqrt{(2 - (-1))^2 + (4 - \frac{3}{2})^2} = \sqrt{(3)^2 + (\frac{8 - 3}{2})^2} = \sqrt{9 + (\frac{5}{2})^2} = \sqrt{9 + \frac{25}{4}} = \sqrt{\frac{36 + 25}{4}} = \sqrt{\frac{61}{4}} = \frac{\sqrt{61}}{2} \)

Length of QR:
\( QR = \sqrt{(5 - 2)^2 + (\frac{3}{2} - 4)^2} = \sqrt{(3)^2 + (\frac{3 - 8}{2})^2} = \sqrt{9 + (-\frac{5}{2})^2} = \sqrt{9 + \frac{25}{4}} = \sqrt{\frac{61}{4}} = \frac{\sqrt{61}}{2} \)

Length of RS:
\( RS = \sqrt{(2 - 5)^2 + (-1 - \frac{3}{2})^2} = \sqrt{(-3)^2 + (\frac{-2 - 3}{2})^2} = \sqrt{9 + (-\frac{5}{2})^2} = \sqrt{9 + \frac{25}{4}} = \sqrt{\frac{61}{4}} = \frac{\sqrt{61}}{2} \)

Length of SP:
\( SP = \sqrt{(-1 - 2)^2 + (\frac{3}{2} - (-1))^2} = \sqrt{(-3)^2 + (\frac{3 + 2}{2})^2} = \sqrt{9 + (\frac{5}{2})^2} = \sqrt{9 + \frac{25}{4}} = \sqrt{\frac{61}{4}} = \frac{\sqrt{61}}{2} \)

Since \( PQ = QR = RS = SP = \frac{\sqrt{61}}{2} \), all four sides are equal. This suggests it could be a rhombus or a square. To differentiate, we check the diagonals.

Length of diagonal PR:
\( PR = \sqrt{(5 - (-1))^2 + (\frac{3}{2} - \frac{3}{2})^2} = \sqrt{(6)^2 + (0)^2} = \sqrt{36} = 6 \)

Length of diagonal QS:
\( QS = \sqrt{(2 - 2)^2 + (-1 - 4)^2} = \sqrt{(0)^2 + (-5)^2} = \sqrt{25} = 5 \)

Since \( PR \neq QS \) (6 ≠ 5), the diagonals are not equal. A square has equal diagonals, while a rhombus has unequal diagonals (unless it's also a square).

Therefore, quadrilateral PQRS is a rhombus.
In simple words: We found the middle points of each side of the rectangle. Then we measured the length of all four sides of the new shape, PQRS, and found they were all equal. This meant it was either a square or a rhombus. Next, we measured the two diagonal lines inside PQRS and saw they were not the same length. Since all sides are equal but the diagonals are not, the shape must be a rhombus.

Exam Tip: To classify a quadrilateral, first calculate the lengths of all four sides. If they are equal, it's a rhombus or a square. Then, calculate the lengths of the diagonals. If the diagonals are also equal, it's a square; if they are unequal, it's a rhombus.