CBSE Class 10 Mathematics Coordinate Geometry Worksheet Set 06

Question 1. Area of triangle enclosed between the coordinate axes with vertices (6, 0) and (0, 11) is _____
(a) 4 sq units
(b) 5 sq units
(c) 10 sq units
(d) None of the options
Answer: (d) None of the options
The triangle is formed by the origin \( (0, 0) \), \( (6, 0) \), and \( (0, 11) \). Since the coordinate axes are perpendicular, this is a right-angled triangle where the base is 6 units and the height is 11 units.
The area is calculated as:
Area \( = \frac{1}{2} \times \text{base} \times \text{height} \)

\( \text{Area} = \frac{1}{2} \times 6 \times 11 = 33 \) sq units.
Since 33 sq units is not among the given options, the correct choice is (d).
In simple words: The three corners of the triangle are at the origin, 6 units along the horizontal axis, and 11 units up the vertical axis. The area is half of 6 times 11, which gives 33 square units.

Exam Tip: In questions where the coordinate axes enclose the triangle, always remember that the origin \( (0, 0) \) is the third vertex, which makes the base and height equal to the non-zero coordinates.

Question 2. Distance between two points A(x1, y1) B(x2, y2) is AB =
(a) \( \sqrt{(x_1 - y_1)^2 + (x_2 - y_2)^2} \)
(b) \( \sqrt{(x_1 - y_1)^2 + (x_2 - y_2)^2} \)
(c) \( \sqrt{(x_1 - y_2)^2 + (y_1 - x_2)^2} \)
(d) \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
Answer: (d) \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
By using the coordinates of any two points, we calculate the straight-line distance using the distance formula, which is \( AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). This is derived directly from the Pythagorean theorem.
In simple words: To find the distance between any two points, subtract their x-coordinates, square the result, do the same for the y-coordinates, add them together, and then find the square root.

Exam Tip: The order of subtraction inside the squares does not affect the final value because the square of a negative number is always positive, but keeping consistent ordering helps prevent minor mistakes.

Question 3. In what ratio is the segment joining the points (5, 1) and (-7, -1) divided by X-axis ?
(a) 1 : 6
(b) 6 : 2
(c) 2 : 6
(d) 1 : 1
Answer: (d) 1 : 1
Let the ratio in which the X-axis divides the line segment be \( k : 1 \). Any point lying on the X-axis has a y-coordinate equal to 0.
Using the section formula for the y-coordinate:
\( y = \frac{k y_2 + 1 y_1}{k + 1} \)

\( 0 = \frac{k(-1) + 1(1)}{k + 1} \)

\( 0 = -k + 1 \)

\( k = 1 \)
Therefore, the ratio is \( 1 : 1 \).
In simple words: Since the X-axis divides the line segment, the dividing point has a height of 0. Calculating with the given heights of 1 and -1 shows that the point is exactly in the middle, splitting the segment into two equal parts.

Exam Tip: Whenever a line segment is divided by the X-axis, start your solution by setting the y-coordinate of the dividing point to zero, as this is the quickest way to find the ratio.

Question 4. The coordinates of the points of trisection of a segment joining A(-3, 2) and B(9, 5) is _____
(a) (3, 1), (-5, -4)
(b) (5, 9), (-9, 5)
(c) (2, 3), (4, 5)
(d) (1, 3), (5, 4)
Answer: (d) (1, 3), (5, 4)
Let the points of trisection be \( P \) and \( Q \). The point \( P \) splits the segment \( AB \) in the ratio \( 1 : 2 \), and the point \( Q \) splits it in the ratio \( 2 : 1 \).
Applying the section formula for \( P \):
\( x_P = \frac{1(9) + 2(-3)}{1 + 2} = \frac{9 - 6}{3} = 1 \)
\( y_P = \frac{1(5) + 2(2)}{1 + 2} = \frac{5 + 4}{3} = 3 \)
Thus, \( P \) is \( (1, 3) \).
Applying the section formula for \( Q \):
\( x_Q = \frac{2(9) + 1(-3)}{2 + 1} = \frac{18 - 3}{3} = 5 \)
\( y_Q = \frac{2(5) + 1(2)}{2 + 1} = \frac{10 + 2}{3} = 4 \)
Thus, \( Q \) is \( (5, 4) \).
In simple words: Trisection means cutting the line into three equal pieces. We find the coordinates of these two cuts by using section ratios of 1 to 2 and 2 to 1, giving us the points (1, 3) and (5, 4).

Exam Tip: Remember that points of trisection always divide a segment into parts in the ratios of \( 1 : 2 \) and \( 2 : 1 \). Do not confuse this with midpoint calculations.

Question 5. If A (3, 1), B(2, 6) and C(-5, 7) are the midpoints of the sides of \( \triangle PQR \), the area of the triangle is
(a) 68 sq. units
(b) 24 sq. units
(c) 48 sq. units
(d) 50 sq. units
Answer: (a) 68 sq. units
The area of a triangle formed by connecting the midpoints of the sides of a larger triangle is always equal to one-fourth of the total area of that larger triangle. Therefore, the area of \( \triangle PQR \) is \( 4 \times \text{Area of } \triangle ABC \).
Let us calculate the area of \( \triangle ABC \) first:
\( \text{Area of } \triangle ABC = \frac{1}{2} | 3(6 - 7) + 2(7 - 1) + (-5)(1 - 6) | \)
\( = \frac{1}{2} | 3(-1) + 2(6) - 5(-5) | \)
\( = \frac{1}{2} | -3 + 12 + 25 | = \frac{1}{2} | 34 | = 17 \) sq. units.
Now, the area of the larger triangle \( \triangle PQR \) is:
\( \text{Area of } \triangle PQR = 4 \times 17 = 68 \) sq. units.
In simple words: The triangle made from the middle points of the sides has an area of 17. Since the main triangle is exactly four times larger than this inner triangle, its area is 68 square units.

Exam Tip: Keep the geometric property in mind - joining the midpoints of any triangle divides it into four congruent triangles of equal area.

Question 6. The coordinates of the point P(x, y) which divides the line segment joining A(x1,y1) and B(x2, y2) internally in the ratio m : n is
(a) \( (x, y) = \left( \frac{m x_1 - n y_1}{m + n}, \frac{m x_2 + n y_2}{m + n} \right) \)
(b) \( (x, y) = \left( \frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n} \right) \)
(c) \( (x, y) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \)
(d) None of the options
Answer: (b) \( (x, y) = \left( \frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n} \right) \)
The internal section formula determines the coordinates of a point dividing a line segment in a specified ratio \( m : n \). The correct formula is \( \left( \frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n} \right) \).
In simple words: The formula tells us where a point lies on a line when divided in a given ratio. We cross-multiply the ratio values with the opposite coordinates and divide by the total ratio sum.

Exam Tip: Be careful with the cross-multiplication in the formula - the ratio value \( m \) must be multiplied by the coordinates of the second point \( B \), and \( n \) by the first point \( A \).

Question 7. A(-1, 2), B(4, 1) and C(7, 6) are three vertices of the parallelogram ABCD. The coordinates of fourth vertex is.
(a) (7, 2)
(b) (-2, 7)
(c) (7, -2)
(d) (2, 7)
Answer: (d) (2, 7)
Let the coordinates of the fourth vertex be \( D(x, y) \). In a parallelogram, the diagonals bisect each other, which means the midpoint of diagonal \( AC \) is the same as the midpoint of diagonal \( BD \).
Midpoint of \( AC \):
\( \left( \frac{-1 + 7}{2}, \frac{2 + 6}{2} \right) = (3, 4) \)
Midpoint of \( BD \):
\( \left( \frac{4 + x}{2}, \frac{1 + y}{2} \right) \)
Equating the coordinates:
\( \frac{4 + x}{2} = 3 \)

\( 4 + x = 6 \)

\( x = 2 \)
And,
\( \frac{1 + y}{2} = 4 \)

\( 1 + y = 8 \)

\( y = 7 \)
So, the coordinates of the fourth vertex are \( (2, 7) \).
In simple words: Since the diagonals of a parallelogram meet exactly in the middle, the midpoint of the line joining the first and third corners must match the midpoint of the line joining the second and fourth corners. Solving this gives the fourth point as (2, 7).

Exam Tip: Using the midpoint property of diagonals is the most efficient method to find the missing vertex of a parallelogram, compared to using the distance formula or slope formula.

Question 8. Area of quadrilateral formed by the vertices (-1, 7), (-4, -9), (9, -3) and (8, -5)
(a) 105 sq. units
(b) 18 sq. units
(c) 50 sq. units
(d) 25 sq. units
Answer: (a) 105 sq. units
Let the coordinates of the vertices be \( A(-1, 7) \), \( B(-4, -9) \), \( C(9, -3) \), and \( D(8, -5) \). To find the area of the non-self-intersecting quadrilateral formed by these vertices, we arrange them in cyclic order \( A(-1, 7) \), \( B(-4, -9) \), \( D(8, -5) \), and \( C(9, -3) \).
We can split this quadrilateral into two triangles: \( \triangle ABD \) and \( \triangle ACD \).
Area of \( \triangle ABD \):
\( \text{Area}_1 = \frac{1}{2} | -1(-9 - (-5)) + (-4)(-5 - 7) + 8(7 - (-9)) | \)
\( = \frac{1}{2} | 4 + 48 + 128 | = \frac{1}{2} | 180 | = 90 \) sq. units.
Area of \( \triangle ACD \):
\( \text{Area}_2 = \frac{1}{2} | -1(-3 - (-5)) + 9(-5 - 7) + 8(7 - (-3)) | \)
\( = \frac{1}{2} | -2 - 108 + 80 | = \frac{1}{2} | -30 | = 15 \) sq. units.
Total Area \( = \text{Area}_1 + \text{Area}_2 = 90 + 15 = 105 \) sq. units.
In simple words: We find the area of the shape by cutting it into two triangles, calculating the space inside each triangle using their corner coordinates, and then adding those two areas together to get 105 square units.

Exam Tip: Be careful with the cyclic order of the points when dividing a quadrilateral into triangles. Plotting the points roughly on a graph helps in selecting the correct diagonal to divide the shape.

Question 9. Find the equation of line from which points (7,1) and (3,5) are equidistant.
Answer: Let the point equidistant from the points \( A(7, 1) \) and \( B(3, 5) \) be \( P(x, y) \). Since \( P \) is equidistant from \( A \) and \( B \), we have \( PA = PB \), which means \( PA^2 = PB^2 \).
Using the distance formula:
\( (x - 7)^2 + (y - 1)^2 = (x - 3)^2 + (y - 5)^2 \)

\( x^2 - 14x + 49 + y^2 - 2y + 1 = x^2 - 6x + 9 + y^2 - 10y + 25 \)
Cancel out \( x^2 \) and \( y^2 \) from both sides:
\( -14x - 2y + 50 = -6x - 10y + 34 \)

\( -8x + 8y + 16 = 0 \)
Divide the entire equation by \( -8 \):
\( x - y - 2 = 0 \)
This is the required equation of the line.
In simple words: The set of points that are at the exact same distance from both given points forms a straight line. We find this line by setting the squared distances equal and simplifying, which gives the equation \( x - y = 2 \).

Exam Tip: This path is the perpendicular bisector of the line segment connecting the two points. You can also solve this by finding the midpoint of the segment and using the negative reciprocal of its slope.

Question 10. The point (9, 2) is at a distance of _________ units from Y-axis.
(a) 2 units
(b) 9 units
(c) 5 units
(d) None of the options
Answer: (b) 9 units
The perpendicular distance of any point \( (x, y) \) from the Y-axis is given by the absolute value of its x-coordinate, which is \( |x| \). For the point \( (9, 2) \), the x-coordinate is 9, so its distance from the Y-axis is 9 units.
In simple words: The x-coordinate tells us how far a point is from the vertical Y-axis. Since the x-coordinate of the point (9, 2) is 9, it is exactly 9 units away.

Exam Tip: Do not confuse the distance from the Y-axis with the distance from the X-axis. The distance from the Y-axis is always the x-coordinate, whereas the distance from the X-axis is the y-coordinate.

Question 11. Find the area of a triangle whose vertices are (3, 8), (-4, 2) and (5, -1).
(a) 35.1 sq unit
(b) 37.5 sq unit
(c) 38 sq unit
(d) 40 sq unit
Answer: (b) 37.5 sq unit
Let the vertices of the triangle be \( (x_1, y_1) = (3, 8) \), \( (x_2, y_2) = (-4, 2) \), and \( (x_3, y_3) = (5, -1) \).
The area of the triangle is calculated as:
Area \( = \frac{1}{2} | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) | \)

\( \text{Area} = \frac{1}{2} | 3(2 - (-1)) + (-4)(-1 - 8) + 5(8 - 2) | \)

\( \text{Area} = \frac{1}{2} | 3(3) - 4(-9) + 5(6) | \)

\( \text{Area} = \frac{1}{2} | 9 + 36 + 30 | \)

\( \text{Area} = \frac{1}{2} | 75 | = 37.5 \) sq units.
In simple words: Using the coordinate formula for the space inside a triangle, we substitute our point values to compute the total area, which works out to 37.5 square units.

Exam Tip: Always use absolute value bars when calculating the area of a triangle because area is a physical quantity that can never be negative.

Question 12. Find the coordinates of centre of circle if the coordinate of end points of its diameter are \( P(x_1, y_1) \) and \( Q(x_2, y_2) \)
Answer: The centre of any circle lies exactly at the middle of its diameter. Therefore, the coordinates of the centre are found using the midpoint formula:
Coordinates of the centre \( = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \).
In simple words: The center of a circle is the middle point of its diameter. We find it by taking the average of the coordinates of both endpoints.

Exam Tip: The midpoint formula is very simple, but make sure to add the coordinates instead of subtracting them, which is a common formula confusion.

Question 13. The point (-3, -2) belongs to Quadrant _____
(a) \( Q_1 \)
(b) \( Q_2 \)
(c) \( Q_3 \)
(d) \( Q_4 \)
Answer: (c) \( Q_3 \)
In a coordinate plane, the signs of the coordinates determine the quadrant:
- Quadrant I: both coordinates are positive \( (+, +) \)
- Quadrant II: x is negative, y is positive \( (-, +) \)
- Quadrant III: both coordinates are negative \( (-, -) \)
- Quadrant IV: x is positive, y is negative \( (+, -) \)
Since both coordinates in \( (-3, -2) \) are negative, the point lies in the third quadrant \( (Q_3) \).
In simple words: Since both numbers in (-3, -2) are negative, we must move left and then down from the center, which places the point in the bottom-left area, known as the third quadrant.

Exam Tip: Memorize the coordinate signs of all four quadrants to quickly identify where any given point is located on the graph.

Question 14. The point (1,-3),(13,9),(10,12) and (-2,0) taken in order form the vertices of__________
(a) square
(b) rectangle
(c) rhombus
(d) none
Answer: (b) rectangle
Let the given points in order be \( A(1, -3) \), \( B(13, 9) \), \( C(10, 12) \), and \( D(-2, 0) \). We find the lengths of the sides using the distance formula:
\( AB = \sqrt{(13 - 1)^2 + (9 - (-3))^2} = \sqrt{12^2 + 12^2} = \sqrt{288} \)
\( BC = \sqrt{(10 - 13)^2 + (12 - 9)^2} = \sqrt{(-3)^2 + 3^2} = \sqrt{18} \)
\( CD = \sqrt{(-2 - 10)^2 + (0 - 12)^2} = \sqrt{(-12)^2 + (-12)^2} = \sqrt{288} \)
\( DA = \sqrt{(1 - (-2))^2 + (-3 - 0)^2} = \sqrt{3^2 + (-3)^2} = \sqrt{18} \)
Since the opposite sides are equal (\( AB = CD \) and \( BC = DA \)), the figure is a parallelogram. Now, we check the lengths of the diagonals \( AC \) and \( BD \):
\( AC = \sqrt{(10 - 1)^2 + (12 - (-3))^2} = \sqrt{9^2 + 15^2} = \sqrt{306} \)
\( BD = \sqrt{(-2 - 13)^2 + (0 - 9)^2} = \sqrt{(-15)^2 + (-9)^2} = \sqrt{306} \)
Since the diagonals are also equal (\( AC = BD \)), the parallelogram is a rectangle.
In simple words: By measuring the distance between the points, we see that opposite sides are equal, which makes it a parallelogram. Since the two diagonal lines connecting opposite corners are also equal in length, the shape is a rectangle.

Exam Tip: A parallelogram becomes a rectangle if its diagonals are equal. Always check both side lengths and diagonal lengths to confirm the geometric shape.

Question 15. Find the distance of a point P(x,y) from origin.
Answer: The coordinates of the origin are \( (0, 0) \). Using the distance formula between \( (0, 0) \) and \( P(x, y) \):
Distance \( = \sqrt{(x - 0)^2 + (y - 0)^2} = \sqrt{x^2 + y^2} \).
In simple words: To find how far any point is from the starting center point (0, 0), square both coordinates, add them, and find the square root.

Exam Tip: This formula is a simplified form of the distance formula. You can write it directly whenever a point's distance from the origin is asked.

Question 16. Find the relation between x and y if the points (x,y),(1,2) and (7,0) are collinear.
Answer: If the points \( (x, y) \), \( (1, 2) \), and \( (7, 0) \) are collinear, the area of the triangle formed by them must be equal to 0.
Using the triangle area formula:
\( \frac{1}{2} | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) | = 0 \)
Substitute the given points:
\( | x(2 - 0) + 1(0 - y) + 7(y - 2) | = 0 \)

\( 2x - y + 7y - 14 = 0 \)

\( 2x + 6y - 14 = 0 \)
Divide the equation by 2:
\( x + 3y - 7 = 0 \)
This is the required relationship between \( x \) and \( y \).
In simple words: Since these three points lie on the exact same straight line, the area of the triangle they form is zero. Simplifying this condition gives us the relation \( x + 3y = 7 \).

Exam Tip: For three points to be collinear, you can also equate the slope of the line joining the first two points with the slope of the line joining the second and third points.

Question 17. X-coordinate is also called as________
Answer: Abscissa.
In simple words: Abscissa is another name for the horizontal x-coordinate of a point.

Exam Tip: Remember these terms as they are frequently used in descriptive questions to refer to coordinates.

Question 18. Y-coordinate is also called as________
Answer: Ordinate.
In simple words: Ordinate is the term used to describe the vertical y-coordinate of a point.

Exam Tip: Always be clear on the distinction between "abscissa" (x-axis distance) and "ordinate" (y-axis distance) to avoid swapping them in calculations.

Question 19. Find the area of the triangle formed by the points A (5, 2), B(4, 7) and C(7, -4).
(a) 10
(b) 6
(c) 2
(d) None of the options
Answer: (c) 2
Let the vertices of the triangle be \( (x_1, y_1) = (5, 2) \), \( (x_2, y_2) = (4, 7) \), and \( (x_3, y_3) = (7, -4) \).
Using the area formula:
Area \( = \frac{1}{2} | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) | \)

\( \text{Area} = \frac{1}{2} | 5(7 - (-4)) + 4(-4 - 2) + 7(2 - 7) | \)

\( \text{Area} = \frac{1}{2} | 5(11) + 4(-6) + 7(-5) | \)

\( \text{Area} = \frac{1}{2} | 55 - 24 - 35 | \)

\( \text{Area} = \frac{1}{2} | -4 | = 2 \) sq. units.
In simple words: Plugging the values of the three points into the area formula tells us that the space enclosed by the triangle is 2 square units.

Exam Tip: When simplifying the expression inside the absolute value, follow the order of operations carefully to avoid sign errors, which are very common in multi-step calculations.

Question 20. Find the coordinates of the point of intersection of two distinct lines if both of them pass through the origin.
Answer: Any line passing through the origin must pass through the point \( (0, 0) \). Since both of these distinct lines pass through the origin, their only common point of intersection is the origin itself, which has the coordinates \( (0, 0) \).
In simple words: Since both lines go through the starting point on the graph, they must cross each other exactly at that point, which is (0, 0).

Exam Tip: This is a conceptual question. No calculation is needed since the origin is, by definition, the common point for all lines of the form \( y = mx \).

Question 21. What will be the area of triangle whose vertices are \( P(x_1, y_1) \), \( Q(x_2, y_2) \) and \( R(x_3, y_3) \)?
Answer: The area of a triangle with given coordinates is calculated using the formula:
Area \( = \frac{1}{2} | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) | \).
In simple words: This formula calculates the total flat space occupied by a triangle when we know the coordinates of its three corners.

Exam Tip: Writing down this general formula correctly is worth step-marks in subjective exams, even if there are arithmetic errors in later calculations.

Question 22. The point (2, 5) is at a distance of ______ units from X-axis.
(a) 2 units
(b) 5 units
(c) 3 units
(d) None of the options
Answer: (b) 5 units
The perpendicular distance of any point \( (x, y) \) from the X-axis is given by the absolute value of its y-coordinate, which is \( |y| \). For the point \( (2, 5) \), the y-coordinate is 5, so its distance from the X-axis is 5 units.
In simple words: The y-coordinate represents how high a point is above the horizontal line. Since the y-coordinate of (2, 5) is 5, it is exactly 5 units away from the X-axis.

Exam Tip: The distance of a point from the X-axis is always equal to its y-coordinate, while the distance from the Y-axis is its x-coordinate. Remembering this simple rule avoids confusion.

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Question 23. If the distance between points (p, -7),(9, -7) is 15 units, then p is______
(a) -3 or 7
(b) -7 or 3
(c) -3 or -7
(d) -6 or 24
Answer: (d) -6 or 24
Using the distance formula:
\( \text{Distance} = \sqrt{(9 - p)^2 + (-7 - (-7))^2} = 15 \)

\( \sqrt{(9 - p)^2 + 0} = 15 \)

\( |9 - p| = 15 \)
This gives us two cases:
1) \( 9 - p = 15 \implies p = -6 \)
2) \( 9 - p = -15 \implies p = 24 \)
So, the value of \( p \) is \( -6 \) or \( 24 \).
In simple words: Since both points have the same vertical position, the distance is simply the horizontal difference between them. This difference must be 15, which means \( p \) can be either 15 units to the left of 9 (which is -6) or 15 units to the right (which is 24).

Exam Tip: When the y-coordinates are identical, you can simplify the distance calculation by working with the absolute difference of the x-coordinates directly, saving valuable exam time.

Question 24. The coordinates of the centre of a circle Which is passing through (1, 2),(3, -4) and (5, -6) is
(a) (2, 11)
(b) (11, 2)
(c) (11, -2)
(d) (-2, 11)
Answer: (b) (11, 2)
Let the coordinates of the centre of the circle be \( C(x, y) \). The distances from the centre to any point on the circle are equal to the radius.
Let \( P(1, 2) \), \( Q(3, -4) \), and \( R(5, -6) \) be the three points. Then \( CP^2 = CQ^2 = CR^2 \).
Using the distance formula:
\( (x - 1)^2 + (y - 2)^2 = (x - 3)^2 + (y + 4)^2 \)

\( x^2 - 2x + 1 + y^2 - 4y + 4 = x^2 - 6x + 9 + y^2 + 8y + 16 \)

\( -2x - 4y + 5 = -6x + 8y + 25 \)

\( 4x - 12y = 20 \implies x - 3y = 5 \) --- (Eq 1)

Also, \( CQ^2 = CR^2 \):
\( (x - 3)^2 + (y + 4)^2 = (x - 5)^2 + (y + 6)^2 \)

\( x^2 - 6x + 9 + y^2 + 8y + 16 = x^2 - 10x + 25 + y^2 + 12y + 36 \)

\( -6x + 8y + 25 = -10x + 12y + 61 \)

\( 4x - 4y = 36 \implies x - y = 9 \) --- (Eq 2)

Subtracting (Eq 1) from (Eq 2):
\( (x - y) - (x - 3y) = 9 - 5 \)

\( 2y = 4 \implies y = 2 \)
Substituting \( y = 2 \) into (Eq 2):
\( x - 2 = 9 \implies x = 11 \).
So, the coordinates of the centre are \( (11, 2) \).
In simple words: Since the distance from the center to all points on the boundary is the same, we write equations representing these equal distances. Solving these equations gives us the coordinates of the center point as (11, 2).

Exam Tip: Be methodical while expanding terms like \( (y + 4)^2 \) and \( (y + 6)^2 \) to prevent sign mistakes, which are the main cause of incorrect answers in these long coordinate equations.

Question 25. The ratio by which P(4, 6) divide the join of A(-2, 3) and B (6, 7) is ________
(a) \( \frac{5}{7} \)
(b) \( \frac{1}{3} \)
(c) \( \frac{2}{3} \)
(d) None of the options
Answer: (d) None of the options
Let the ratio in which \( P(4, 6) \) divides the segment joining \( A(-2, 3) \) and \( B(6, 7) \) be \( k : 1 \).
Using the section formula for the y-coordinate:
\( y_P = \frac{k y_2 + 1 y_1}{k + 1} \)

\( 6 = \frac{k(7) + 1(3)}{k + 1} \)

\( 6k + 6 = 7k + 3 \)

\( k = 3 \)
So, the ratio is \( 3 : 1 \). Since \( 3 : 1 \) is not represented by any of the options (a), (b), or (c), the correct choice is (d).
In simple words: We find the dividing ratio using the section formula on the vertical coordinates. This shows that the point divides the line in a 3 to 1 ratio, which is not listed in the options.

Exam Tip: Always verify your ratio by substituting the value of \( k \) back into the formula for the other coordinate (x-coordinate in this case) to make sure it yields the correct value.

Question 26. Two vertices of triangle are (-1, 4) and (5, 2) and medians intersect at (0, -3). Then the third vertex is
(a) (4, 15)
(b) (3, 15)
(c) (-4, 15)
(d) None of the options
Answer: (d) None of the options
The point of intersection of the medians is the centroid of the triangle, which is \( G(0, -3) \). Let the third vertex be \( C(x, y) \).
Using the centroid formula:
\( G_x = \frac{x_1 + x_2 + x_3}{3} \)

\( 0 = \frac{-1 + 5 + x}{3} \implies 4 + x = 0 \implies x = -4 \)
And,
\( G_y = \frac{y_1 + y_2 + y_3}{3} \)

\( -3 = \frac{4 + 2 + y}{3} \implies 6 + y = -9 \implies y = -15 \)
Thus, the third vertex is \( (-4, -15) \). Since \( (-4, -15) \) is not in the options, the correct choice is (d).
In simple words: The intersection of the medians is the center point (centroid). Using the average formula for the three corners, we find that the third corner must be at (-4, -15), which is not among the given choices.

Exam Tip: Remember that "intersection of medians" is simply another term for the centroid of the triangle. Use the centroid formula directly in such questions.

Question 27. A(-1, 0), B(3, 1) and C(2, 2) are three vertices of the parallelogram ABCD. The coordinates of fourth vertex is
(a) (-2, -1)
(b) (2, 1)
(c) (2, -1)
(d) (-2, 1)
Answer: (d) (-2, 1)
Let the fourth vertex be \( D(x, y) \). In a parallelogram \( ABCD \), the midpoint of diagonal \( AC \) must be the same as the midpoint of diagonal \( BD \).
Midpoint of \( AC \):
\( \left( \frac{-1 + 2}{2}, \frac{0 + 2}{2} \right) = \left( \frac{1}{2}, 1 \right) \)
Midpoint of \( BD \):
\( \left( \frac{3 + x}{2}, \frac{1 + y}{2} \right) \)
Equating the coordinates:
\( \frac{3 + x}{2} = \frac{1}{2} \implies 3 + x = 1 \implies x = -2 \)
And,
\( \frac{1 + y}{2} = 1 \implies 1 + y = 2 \implies y = 1 \)
So, the coordinates of the fourth vertex are \( (-2, 1) \).
In simple words: Parallelogram diagonals cross exactly in their middle. By matching the middle point of the first diagonal line with the second one, we find that the fourth corner must be at (-2, 1).

Exam Tip: Make sure to match the correct pairs of vertices to form the diagonals. For a parallelogram \( ABCD \), the diagonals are always \( AC \) and \( BD \).

Question 28. The point of intersection of x and y axes is called
(a) origin
(b) null point
(c) common point
(d) None of the options
Answer: (a) origin
The horizontal axis (x-axis) and the vertical axis (y-axis) cross each other at a specific starting point with coordinates \( (0, 0) \). This point is universally referred to as the origin.
In simple words: The exact spot where the horizontal and vertical lines cross each other is called the origin, and its coordinate values are always zero.

Exam Tip: The coordinates of the origin are always \( (0, 0) \). This is a standard coordinate geometry concept that you should know by heart.

Question 29. The point on X-axis equidistant from (5, 4) and (-1, 2) is
(a) (3, 0)
(b) (2, 0)
(c) (4, 0)
(d) None of the options
Answer: (a) (3, 0)
Let the point on the X-axis be \( P(x, 0) \). Since \( P \) is equidistant from \( A(5, 4) \) and \( B(-1, 2) \), we have \( PA = PB \), which means \( PA^2 = PB^2 \).
Using the distance formula:
\( (x - 5)^2 + (0 - 4)^2 = (x - (-1))^2 + (0 - 2)^2 \)

\( x^2 - 10x + 25 + 16 = x^2 + 2x + 1 + 4 \)

\( -10x + 41 = 2x + 5 \)

\( 12x = 36 \implies x = 3 \)
So, the required point is \( (3, 0) \).
In simple words: Since the point lies on the X-axis, its height is 0. Setting its distance to both given points as equal and solving the equation gives us the point (3, 0).

Exam Tip: Any point on the X-axis has its y-coordinate as 0. This single fact simplifies the problem by leaving only one variable to solve.

Question 30. The coordinates of the centroid of a triangle whose vertices are (-1, 3),(-4, 5) and (8, 13) is _____
(a) (6, 6)
(b) (6, 0)
(c) (1, 7)
(d) None of the options
Answer: (c) (1, 7)
Let the vertices of the triangle be \( (-1, 3) \), \( (-4, 5) \), and \( (8, 13) \).
Using the centroid formula:
Centroid \( = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \)

\( \text{Centroid} = \left( \frac{-1 + (-4) + 8}{3}, \frac{3 + 5 + 13}{3} \right) \)

\( \text{Centroid} = \left( \frac{3}{3}, \frac{21}{3} \right) = (1, 7) \).
In simple words: We find the center point of the triangle by taking the average of all three x-coordinates and the average of all three y-coordinates, giving us (1, 7).

Exam Tip: Centroid calculation is highly scoring because it only involves simple averages. Double-check your addition of positive and negative numbers to ensure accuracy.

Question 31. The value of K if (-3, 12), (7, 6) and (K, 9) are collinear is
(a) 3
(b) 4
(c) 2
(d) 1
Answer: (c) 2
Since the points are collinear, the area of the triangle formed by them must be 0.
Using the area formula:
\( \frac{1}{2} | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) | = 0 \)
Substitute the points:
\( | -3(6 - 9) + 7(9 - 12) + K(12 - 6) | = 0 \)

\( | -3(-3) + 7(-3) + 6K | = 0 \)

\( | 9 - 21 + 6K | = 0 \)

\( -12 + 6K = 0 \)

\( 6K = 12 \implies K = 2 \).
In simple words: Since these points are on a straight line, they cannot enclose any area. Setting the triangle area formula to zero and solving for the unknown value gives us \( K = 2 \).

Exam Tip: This type of question can also be solved by showing that the slope between the first two points is equal to the slope between the second and third points.

Question 32. The points on X-axis at a distance of 10 units from (11, 8) are
(a) (7, 4) (-16, 5)
(b) (6, -9) (17, -11)
(c) (5, 0) (17, 0)
(d) none of these
Answer: (c) (5, 0) (17, 0)
Let the point on the X-axis be \( (x, 0) \).
Using the distance formula:
\( \sqrt{(x - 11)^2 + (0 - 8)^2} = 10 \)
Square both sides:
\( (x - 11)^2 + 64 = 100 \)

\( (x - 11)^2 = 36 \)
Taking the square root on both sides:
\( x - 11 = 6 \implies x = 17 \)
or \( x - 11 = -6 \implies x = 5 \).
So, the points are \( (5, 0) \) and \( (17, 0) \).
In simple words: The points we want lie on the horizontal line, meaning their height is 0. Using the distance formula, we find two possible horizontal positions that are exactly 10 units away, which are 5 and 17.

Exam Tip: Do not forget that taking the square root of a number yields both positive and negative values, which is why we get two distinct points as the answer.

Question 33. The area of a ABC with vertices A(x1, y1), B(x2, y2) and C(x3, y3) is
(a) \( \frac{1}{2} | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) | \)
(b) \( \frac{1}{2} | x_1(y_2 - y_3) - x_2(y_3 - y_1) - x_3(y_1 - y_2) | \)
(c) \( | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) | \)
(d) \( \frac{1}{2} | x_1(y_2 + y_3) - x_2(y_3 + y_1) - x_3(y_1 + y_2) | \)
Answer: (a) \( \frac{1}{2} | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) | \)
The standard coordinate formula to calculate the area of any triangle \( ABC \) is given by \( \frac{1}{2} | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) | \).
In simple words: This is the standard formula used to find the area of a triangle when the coordinates of its three vertices are given.

Exam Tip: Memorize the cyclic order of coordinates in this formula: \( 1 \rightarrow 2 \rightarrow 3 \rightarrow 1 \). Notice how the x-coordinates \( x_1, x_2, x_3 \) multiply the differences of the cyclic y-coordinates.

Question 34. Distance between two points will be
Answer: For any two points \( P(x_1, y_1) \) and \( Q(x_2, y_2) \), the straight-line distance is given by:
\( PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).
In simple words: The distance formula gives the length of the straight line joining any two points on a graph.

Exam Tip: This is the most fundamental formula in coordinate geometry. Keep it on your fingertips for solving complex coordinate problems.

Question 35. Two vertices of triangle are (3, -5) and (-7, 4). If its centroid is (2, -1), the third vertex is
(a) (4, 5)
(b) (10, -2)
(c) (-4, 5)
(d) None of the options
Answer: (b) (10, -2)
Let the third vertex be \( C(x, y) \). The coordinates of the centroid are \( G(2, -1) \).
Using the centroid formula:
\( G_x = \frac{x_1 + x_2 + x_3}{3} \)

\( 2 = \frac{3 - 7 + x}{3} \implies 6 = -4 + x \implies x = 10 \)
And,
\( G_y = \frac{y_1 + y_2 + y_3}{3} \)

\( -1 = \frac{-5 + 4 + y}{3} \implies -3 = -1 + y \implies y = -2 \)
So, the third vertex is \( (10, -2) \).
In simple words: Since the average of all three corners' positions gives the center, we set up equations using this average to solve for the coordinates of the third corner, which are (10, -2).

Exam Tip: Centroid problems are straightforward to solve. Remember that the sum of the coordinates of the three vertices is equal to three times the coordinates of the centroid.

Question 36. Area of quadrilateral formed by the vertices (1, 1), (7, -3), (12, 2) and (7, 21)
(a) 132 sq. units
(b) 128 sq. units
(c) 150 sq. units
(d) 130 sq. units
Answer: (a) 132 sq. units
Let the vertices of the quadrilateral in cyclic order be \( A(1, 1) \), \( B(7, -3) \), \( C(12, 2) \), and \( D(7, 21) \). We can find the area of the quadrilateral \( ABCD \) by splitting it into two triangles, \( \triangle ABC \) and \( \triangle ACD \).
Using the Shoelace formula directly for the quadrilateral:
Area \( = \frac{1}{2} | (x_1 y_2 - y_1 x_2) + (x_2 y_3 - y_2 x_3) + (x_3 y_4 - y_3 x_4) + (x_4 y_1 - y_4 x_1) | \)

\( \text{Area} = \frac{1}{2} | (1(-3) - 1(7)) + (7(2) - (-3)(12)) + (12(21) - 2(7)) + (7(1) - 21(1)) | \)

\( \text{Area} = \frac{1}{2} | (-3 - 7) + (14 + 36) + (252 - 14) + (7 - 21) | \)

\( \text{Area} = \frac{1}{2} | -10 + 50 + 238 - 14 | \)

\( \text{Area} = \frac{1}{2} | 264 | = 132 \) sq. units.
In simple words: By using the coordinate formula for the overall shape, we calculate the absolute sum of the crossed coordinates, giving us a final area of 132 square units.

Exam Tip: The Shoelace formula is an incredibly fast way to find the area of polygons on coordinate grids. Ensure the vertices are entered in order (clockwise or counter-clockwise).

Question 37. Is (0,0)the mid point of (0, 1) and (0,-1)?
Answer: Yes, it is the midpoint. Let us verify this by calculating the midpoint of the points \( A(0, 1) \) and \( B(0, -1) \):
Midpoint \( = \left( \frac{0 + 0}{2}, \frac{1 + (-1)}{2} \right) = (0, 0) \).
Therefore, \( (0, 0) \) is indeed the midpoint of the given points.
In simple words: Yes, because when we find the exact middle of the horizontal values and the vertical values, we get (0, 0).

Exam Tip: Midpoint questions require simple calculations. If you find the coordinates are opposite values (like 1 and -1), their sum is 0, making the midpoint 0.

Question 38. Find the distance between the points P(-6, 7) and Q(-1, -5)
(a) 11
(b) 13
(c) 15
(d) 20
Answer: (b) 13
Using the distance formula:
\( PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

\( PQ = \sqrt{(-1 - (-6))^2 + (-5 - 7)^2} \)

\( PQ = \sqrt{5^2 + (-12)^2} \)

\( PQ = \sqrt{25 + 144} = \sqrt{169} = 13 \).
In simple words: The distance between the two points is calculated by finding the horizontal change (5 units) and vertical change (12 units). Using Pythagoras' theorem on these values gives us a straight-line distance of 13 units.

Exam Tip: Notice that the side lengths 5, 12, and 13 form a Pythagorean triplet. Recognizing these triplets helps in verifying your calculations quickly.

Question 39. The point (0, -2) lies on
(a) +ve X-axis
(b) +ve Y-axis
(c) -ve X-axis
(d) -ve Y-axis
Answer: (d) -ve Y-axis
For any point \( (x, y) \), if the x-coordinate is 0, the point must lie on the Y-axis. Since the y-coordinate is negative \( (-2) \), the point lies on the negative side of the Y-axis.
In simple words: Because the horizontal position is 0, the point doesn't move left or right. The vertical value is -2, meaning we move straight down, which puts the point on the negative Y-axis.

Exam Tip: Remember that any point with a zero x-coordinate lies on the Y-axis, while any point with a zero y-coordinate lies on the X-axis.

Question 40. The vertices of a triangle are (2, 1), (5, 2) and (3, 4). Find the co-ordinates of the centroid.
(a) \( \left( \frac{10}{3}, \frac{7}{3} \right) \)
(b) \( \left( \frac{5}{3}, \frac{2}{3} \right) \)
(c) \( \left( \frac{7}{3}, \frac{5}{3} \right) \)
(d) \( \left( \frac{7}{3}, \frac{-7}{3} \right) \)
Answer: (a) \( \left( \frac{10}{3}, \frac{7}{3} \right) \)
Let the vertices of the triangle be \( (2, 1) \), \( (5, 2) \), and \( (3, 4) \).
Using the centroid formula:
Centroid \( = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \)

\( \text{Centroid} = \left( \frac{2 + 5 + 3}{3}, \frac{1 + 2 + 4}{3} \right) = \left( \frac{10}{3}, \frac{7}{3} \right) \).
In simple words: The center coordinate is the average of the coordinates of all three corners, which is \( \left( \frac{10}{3}, \frac{7}{3} \right) \).

Exam Tip: Centroid coordinates are often fractions. Do not try to convert them to decimal numbers unless explicitly asked in the question.

Question 41. Three vertices of a rectangle are the points (3,4), (-1,2) and (2,-4); what are the co-ordinates of the fourth vertex?
(a) (1, 4)
(b) (3, 2)
(c) (4, -4)
(d) (6, -2)
Answer: (d) (6, -2)
Let the vertices of the rectangle in cyclic order be \( A(3, 4) \), \( B(-1, 2) \), and \( C(2, -4) \). Let the coordinates of the fourth vertex be \( D(x, y) \). Since a rectangle is a type of parallelogram, its diagonals bisect each other. This means the midpoint of \( AC \) is identical to the midpoint of \( BD \).
Midpoint of \( AC \):
\( \left( \frac{3 + 2}{2}, \frac{4 + (-4)}{2} \right) = \left( \frac{5}{2}, 0 \right) \)
Midpoint of \( BD \):
\( \left( \frac{-1 + x}{2}, \frac{2 + y}{2} \right) \)
Equating both midpoints:
\( \frac{-1 + x}{2} = \frac{5}{2} \implies -1 + x = 5 \implies x = 6 \)
And,
\( \frac{2 + y}{2} = 0 \implies 2 + y = 0 \implies y = -2 \).
Therefore, the coordinates of the fourth vertex are \( (6, -2) \).
In simple words: Since the diagonals of a rectangle meet exactly in the middle, we match the middle point of the diagonal line \( AC \) with the middle of diagonal line \( BD \) to find the coordinates of the fourth corner, which are (6, -2).

Exam Tip: You can quickly verify your answer by ensuring that the dot product of adjacent vectors (like \( \vec{AB} \) and \( \vec{BC} \)) is zero, confirming the \( 90^\circ \) angles of the rectangle.

Question 42. If points A(x1, y1), B(x2, y2) and C(x3, y3) are collinear, then
(a) \( | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) | = 0 \)
(b) \( | x_1(y_2 - y_3) - x_2(y_3 - y_1) - x_3(y_1 - y_2) | = 0 \)
(c) None of the options
(d) \( | x_1(y_2 + y_3) - x_2(y_3 + y_1) - x_3(y_1 + y_2) | = 0 \)
Answer: (a) \( | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) | = 0 \)
When three points are collinear, they lie on a single line and do not form a triangle. Consequently, the area of the triangle with these points as vertices is 0. Using the area formula, this gives the condition \( | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) | = 0 \).
In simple words: Points that are on the same line cannot form a triangle, meaning the mathematical area formula for their triangle must equal zero.

Exam Tip: Collinearity can be solved using either the area formula or by equating slopes. Use the area formula when coordinates contain variables that are easier to solve linearly.

Question 43. Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).
(a) 0, -7
(b) -7, 0
(c) 7, -7
(d) -7, 7
Answer: (b) -7, 0
Let the point on the x-axis be \( P(x, 0) \). This point is at the same distance from \( A(2, -5) \) and \( B(-2, 9) \), meaning \( PA^2 = PB^2 \).
Using the distance formula:
\( (x - 2)^2 + (0 - (-5))^2 = (x - (-2))^2 + (0 - 9)^2 \)

\( x^2 - 4x + 4 + 25 = x^2 + 4x + 4 + 81 \)
Subtracting \( x^2 \) and \( 4 \) from both sides:
\( -4x + 25 = 4x + 81 \)

\( 8x = -56 \implies x = -7 \).
So, the required point is \( (-7, 0) \).
In simple words: A point on the x-axis has a y-value of 0. Setting up the equation where the distance from this point to both given points is equal gives us the position (-7, 0).

Exam Tip: Be precise when simplifying equations where the squared terms cancel out on both sides, as algebraic sign errors are common in these calculations.

Question 44. Find the co-ordinates of the point which divides the line joining the points (-1, 7) and (4, -3) in the ratio 2 : 3.
(a) (2, 3)
(b) (3, 3)
(c) (1, 3)
(d) (3, 2)
Answer: (c) (1, 3)
Let the dividing point be \( P(x, y) \), which divides the segment joining \( A(-1, 7) \) and \( B(4, -3) \) in the ratio \( m : n = 2 : 3 \).
Using the internal section formula:
\( x = \frac{m x_2 + n x_1}{m + n} \)

\( x = \frac{2(4) + 3(-1)}{2 + 3} = \frac{8 - 3}{5} = 1 \)
And,
\( y = \frac{m y_2 + n y_1}{m + n} \)

\( y = \frac{2(-3) + 3(7)}{2 + 3} = \frac{-6 + 21}{5} = 3 \).
So, the coordinates are \( (1, 3) \).
In simple words: The section formula helps us find a point that is two-fifths of the way along the line from the first point to the second point. Calculating this gives the coordinate values (1, 3).

Exam Tip: Always write down the section formula clearly before plugging in the values, as this ensures partial credit even if there is a calculation error.

Question 45. Find the area of the triangle whose vertices are (3, 2), (11, 8) and (8, 12).
(a) 21
(b) 25
(c) 23
(d) None of the options
Answer: (b) 25
Let the vertices of the triangle be \( (x_1, y_1) = (3, 2) \), \( (x_2, y_2) = (11, 8) \), and \( (x_3, y_3) = (8, 12) \).
The area of the triangle is:
Area \( = \frac{1}{2} | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) | \)

\( \text{Area} = \frac{1}{2} | 3(8 - 12) + 11(12 - 2) + 8(2 - 8) | \)

\( \text{Area} = \frac{1}{2} | 3(-4) + 11(10) + 8(-6) | \)

\( \text{Area} = \frac{1}{2} | -12 + 110 - 48 | \)

\( \text{Area} = \frac{1}{2} | 50 | = 25 \) sq. units.
In simple words: We find the area of the triangle by using the coordinate formula, which gives us a final space of 25 square units.

Exam Tip: Be sure to keep track of the signs when multiplying. A single sign error during the summation stage can completely change the calculated area.

Question 46. Find the distance of the point (6, -6) from the origin.
(a) 2
(b) 6
(c) \( 6\sqrt{2} \)
(d) 0
Answer: (c) \( 6\sqrt{2} \)
The distance of any point \( (x, y) \) from the origin \( (0, 0) \) is given by the formula \( \sqrt{x^2 + y^2} \).
For the point \( (6, -6) \):
Distance \( = \sqrt{6^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2} \).
In simple words: To find how far this point is from the center, we square both coordinates, add them to get 72, and take the square root, which simplifies to \( 6\sqrt{2} \) units.

Exam Tip: Remember to simplify the square root completely by finding the largest square factor (in this case, 36) to write the final answer in surd form.

Question 47. For what value of x will the points (x, 1), (2, 1) and (4, 5) lie on a line?
(a) 1
(b) -1
(c) 2
(d) -2
Answer: (c) 2
If the points are on the same line, they are collinear, which means the area of the triangle formed by them must be 0.
Using the area formula:
\( \frac{1}{2} | x(1 - 5) + 2(5 - 1) + 4(1 - 1) | = 0 \)

\( | x(-4) + 2(4) + 4(0) | = 0 \)

\( -4x + 8 = 0 \)

\( 4x = 8 \implies x = 2 \).
In simple words: Since these points are on a straight line, they cannot enclose any space. Setting the area formula of the triangle to zero and solving for \( x \) gives us the value 2.

Exam Tip: Collinear questions often contain simple coordinate values where terms become zero, simplifying the algebra significantly. Look for these zeroes to speed up your calculations.

Question 48. Find the coordinates of the centroid of a triangle whose vertices are (-2, 4), (-5, 6) and (13, 14).
(a) (5, 5)
(b) (7, 0)
(c) (2, 8)
(d) None of the options
Answer: (c) (2, 8)
Let the vertices of the triangle be \( (x_1, y_1) = (-2, 4) \), \( (x_2, y_2) = (-5, 6) \), and \( (x_3, y_3) = (13, 14) \).
Using the centroid formula:
Centroid \( = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \)

\( \text{Centroid} = \left( \frac{-2 + (-5) + 13}{3}, \frac{4 + 6 + 14}{3} \right) = \left( \frac{6}{3}, \frac{24}{3} \right) = (2, 8) \).
In simple words: The center of the triangle is found by averaging the horizontal positions and vertical positions of all three corners, which gives the point (2, 8).

Exam Tip: Centroid calculation is simple arithmetic. Be careful with calculations involving negative coordinates like \( -2 \) and \( -5 \).

Question 49. Find the value of K if P(K, 10) is the mid point of AB, where the co-ordinate of A and B are (-4, 13) and (8, 7) respectively.
(a) 3
(b) 4
(c) 2
(d) 1
Answer: (c) 2
Since \( P(K, 10) \) is the midpoint of the line segment \( AB \), we apply the midpoint formula for the x-coordinate:
\( K = \frac{x_1 + x_2}{2} \)
Substitute the values of the x-coordinates of \( A \) and \( B \):
\( K = \frac{-4 + 8}{2} = \frac{4}{2} = 2 \).
Thus, the value of \( K \) is 2.
In simple words: The midpoint's horizontal position \( K \) must be the average of the starting and ending horizontal positions. The average of -4 and 8 is 2, so \( K = 2 \).

Exam Tip: Midpoint questions are highly scoring. You do not need to calculate both coordinates if only one variable in a single coordinate is asked.

Question 50. Area of triangle enclosed between the coordinate axes with vertices (5, 0) and (0, 8) is _____
(a) 3 sq units
(b) 5 sq units
(c) 8 sq units
(d) None of the options
Answer: (d) None of the options
The triangle is enclosed between the coordinate axes, which means its third vertex is the origin \( (0, 0) \). This forms a right-angled triangle where the base is 5 units (along the X-axis) and the height is 8 units (along the Y-axis).
The area is:
Area \( = \frac{1}{2} \times \text{base} \times \text{height} \)

\( \text{Area} = \frac{1}{2} \times 5 \times 8 = 20 \) sq units.
Since 20 sq units is not among the options (a), (b), or (c), the correct choice is (d).
In simple words: The triangle's corners are at the origin (0, 0), and along the axes at distances of 5 and 8. The area is half of 5 times 8, which is 20 square units.

Exam Tip: When a triangle is enclosed by the axes, the non-zero coordinate values of the intercepts are directly the base and height of the right-angled triangle.

Question 51. Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.
(a) collinear
(b) non collinear
(c) Both of the options
(d) None of the options
Answer: (b) non collinear
Let the points be \( A(1, 5) \), \( B(2, 3) \), and \( C(-2, -11) \). We find the area of the triangle formed by these three points:
Area \( = \frac{1}{2} | 1(3 - (-11)) + 2(-11 - 5) + (-2)(5 - 3) | \)

\( \text{Area} = \frac{1}{2} | 1(14) + 2(-16) - 2(2) | \)

\( \text{Area} = \frac{1}{2} | 14 - 32 - 4 | \)

\( \text{Area} = \frac{1}{2} | -22 | = 11 \) sq. units.
Since the area is not equal to 0, the points do not lie on a single straight line and are therefore non-collinear.
In simple words: Since the area calculated using the coordinates is 11 (which is not zero), these three points form a proper triangle and do not lie on a single straight line.

Exam Tip: If the area of a triangle formed by three points is non-zero, they are non-collinear. If the area is exactly zero, they are collinear.

Question 52. If the distance between points (a, -5),(7, -5) is 12 units, then find the value of a.
(a) -5 or 5
(b) -5 or 4
(c) -4 or -5
(d) -5 or 19
Answer: (d) -5 or 19
Using the distance formula:
\( \text{Distance} = \sqrt{(7 - a)^2 + (-5 - (-5))^2} = 12 \)

\( \sqrt{(7 - a)^2 + 0} = 12 \)

\( |7 - a| = 12 \)
This gives two possibilities:
1) \( 7 - a = 12 \implies a = -5 \)
2) \( 7 - a = -12 \implies a = 19 \)
So, the value of \( a \) is \( -5 \) or \( 19 \).
In simple words: Since the vertical height of both points is the same, the distance is just the horizontal difference between them. The difference must be 12, which means \( a \) can be either 12 units to the left of 7 (which is -5) or 12 units to the right (which is 19).

Exam Tip: Absolute value equations have two solutions. Make sure to solve for both cases to get the complete set of values.

Question 53. Find the value of x, if the distance between the points (x, -1) and (3, 2) is 5.
(a) 0, -1
(b) 1, 1
(c) 7, 1
(d) 7, -1
Answer: (d) 7, -1
Using the distance formula:
\( \sqrt{(3 - x)^2 + (2 - (-1))^2} = 5 \)
Square both sides:
\( (3 - x)^2 + 3^2 = 25 \)

\( (3 - x)^2 + 9 = 25 \)

\( (3 - x)^2 = 16 \)
Taking square root on both sides:
\( 3 - x = 4 \implies x = -1 \)
or \( 3 - x = -4 \implies x = 7 \).
So, the value of \( x \) is \( 7 \) or \( -1 \).
In simple words: Setting up the distance equation, we find that the squared horizontal distance plus the squared vertical distance must equal 25. Solving this quadratic relation yields two possible horizontal coordinate values: 7 and -1.

Exam Tip: Be sure to write both options in the final answer when a square root is taken during algebraic simplification.

Question 54. Find the mid-point of the line joining (-3, -6) and (1, -2).
(a) (1, -3)
(b) (-3, -1)
(c) (1, -4)
(d) (-1, -4)
Answer: (d) (-1, -4)
Let the points be \( (-3, -6) \) and \( (1, -2) \).
Using the midpoint formula:
Midpoint \( = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \)

\( \text{Midpoint} = \left( \frac{-3 + 1}{2}, \frac{-6 + (-2)}{2} \right) = \left( \frac{-2}{2}, \frac{-8}{2} \right) = (-1, -4) \).
In simple words: The midpoint is found by taking the average of the two horizontal coordinates and the average of the two vertical coordinates, giving us the point (-1, -4).

Exam Tip: The midpoint formula is simple but watch out for negative signs when adding coordinates together.

Question 55. Do the points (3, 2), (-2, -3) and (2, 3) form a triangle? If so, name the type of triangle formed.
Answer: Let the points be \( A(3, 2) \), \( B(-2, -3) \), and \( C(2, 3) \).
We calculate the lengths of the sides using the distance formula:
\( AB = \sqrt{(-2 - 3)^2 + (-3 - 2)^2} = \sqrt{(-5)^2 + (-5)^2} = \sqrt{50} \approx 7.07 \) units.
\( BC = \sqrt{(2 - (-2))^2 + (3 - (-3))^2} = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} \approx 7.21 \) units.
\( CA = \sqrt{(3 - 2)^2 + (2 - 3)^2} = \sqrt{1^2 + (-1)^2} = \sqrt{2} \approx 1.41 \) units.
Since the sum of any two sides is greater than the third side (e.g., \( AB + CA > BC \)), these points form a triangle.
Now, let us examine the squares of the side lengths:
\( AB^2 = 50 \)
\( BC^2 = 52 \)
\( CA^2 = 2 \)
Since \( AB^2 + CA^2 = 50 + 2 = 52 = BC^2 \), the triangle satisfies the converse of Pythagoras' theorem.
Therefore, \( \triangle ABC \) is a right-angled triangle.
In simple words: Measuring the side lengths shows that they satisfy the triangle inequality, meaning they can form a triangle. Additionally, because the squares of two shorter sides add up to the square of the longest side, it is a right-angled triangle.

Exam Tip: To show that points form a triangle, first confirm that the sum of any two side lengths is strictly greater than the third side length.

Question 56. Show that the points (1, 7), (4, 2), (-1, -1) and (- 4, 4) are the vertices of a square.
Answer: Let the given points be \( A(1, 7) \), \( B(4, 2) \), \( C(-1, -1) \), and \( D(-4, 4) \).
We find the lengths of the four sides using the distance formula:
\( AB = \sqrt{(4 - 1)^2 + (2 - 7)^2} = \sqrt{3^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34} \) units.
\( BC = \sqrt{(-1 - 4)^2 + (-1 - 2)^2} = \sqrt{(-5)^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34} \) units.
\( CD = \sqrt{(-4 - (-1))^2 + (4 - (-1))^2} = \sqrt{(-3)^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34} \) units.
\( DA = \sqrt{(1 - (-4))^2 + (7 - 4)^2} = \sqrt{5^2 + 3^2} = \sqrt{25 + 9} = \sqrt{34} \) units.
Since all four sides are equal (\( AB = BC = CD = DA = \sqrt{34} \)), \( ABCD \) is a rhombus.
Now, we check the lengths of the diagonals \( AC \) and \( BD \):
\( AC = \sqrt{(-1 - 1)^2 + (-1 - 7)^2} = \sqrt{(-2)^2 + (-8)^2} = \sqrt{4 + 64} = \sqrt{68} \) units.
\( BD = \sqrt{(-4 - 4)^2 + (4 - 2)^2} = \sqrt{(-8)^2 + 2^2} = \sqrt{64 + 4} = \sqrt{68} \) units.
Since the sides are equal and the diagonals are also equal (\( AC = BD = \sqrt{68} \)), \( ABCD \) is a square.
In simple words: We calculate the length of all four sides and find they are all equal, meaning the shape is a rhombus. Next, we measure the diagonal distances from opposite corners and find they are also equal, proving the shape is a square.

Exam Tip: Equal sides alone only prove a shape is a rhombus. You must also prove that the diagonals are equal to confirm that it is indeed a square.

Question 57. Find a relation between x and y such that the point (x , y) is equidistant from the points (7, 1) and (3, 5).
Answer: Let the point \( P(x, y) \) be equidistant from the points \( A(7, 1) \) and \( B(3, 5) \). Since \( P \) is equidistant from \( A \) and \( B \), we have \( PA = PB \), which means \( PA^2 = PB^2 \).
Using the distance formula:
\( (x - 7)^2 + (y - 1)^2 = (x - 3)^2 + (y - 5)^2 \)

\( x^2 - 14x + 49 + y^2 - 2y + 1 = x^2 - 6x + 9 + y^2 - 10y + 25 \)
Subtracting \( x^2 \) and \( y^2 \) from both sides:
\( -14x - 2y + 50 = -6x - 10y + 34 \)

\( -8x + 8y + 16 = 0 \)
Dividing by \( -8 \):
\( x - y - 2 = 0 \).
This is the required relationship between \( x \) and \( y \).
In simple words: Since the point is equally far from both coordinates, we write their distance equations, set them equal, and simplify. This leaves us with the relation \( x - y = 2 \).

Exam Tip: Equidistant relation equations always simplify to a linear equation of the form \( ax + by + c = 0 \), representing a straight line.

Question 58. Find a point on the y-axis which is equidistant from the points A(6, 5) and B(- 4, 3).
Answer: Any point on the y-axis has coordinates of the form \( P(0, y) \). Since \( P \) is equidistant from \( A(6, 5) \) and \( B(-4, 3) \), we have \( PA = PB \), which means \( PA^2 = PB^2 \).
Using the distance formula:
\( (0 - 6)^2 + (y - 5)^2 = (0 - (-4))^2 + (y - 3)^2 \)

\( 36 + y^2 - 10y + 25 = 16 + y^2 - 6y + 9 \)
Subtracting \( y^2 \) from both sides:
\( -10y + 61 = -6y + 25 \)

\( 4y = 36 \implies y = 9 \).
So, the required point is \( (0, 9) \).
In simple words: Since the point lies on the y-axis, its horizontal position is 0. Setting up the equation where the distance from this point to both given coordinates is equal tells us that the vertical position is 9, making the point (0, 9).

Exam Tip: Remember that any point on the y-axis has a zero x-coordinate. Substituting \( x = 0 \) immediately simplifies the problem to a single variable.

Question 59. Find the coordinates of the point which divides the line segment joining the points (4, - 3) and (8, 5) in the ratio 3 : 1 internally.
Answer: Let the dividing point be \( P(x, y) \), which divides the segment joining \( A(4, -3) \) and \( B(8, 5) \) in the ratio \( m : n = 3 : 1 \).
Using the internal section formula:
\( x = \frac{m x_2 + n x_1}{m + n} = \frac{3(8) + 1(4)}{3 + 1} = \frac{24 + 4}{4} = 7 \)
And,
\( y = \frac{m y_2 + n y_1}{m + n} = \frac{3(5) + 1(-3)}{3 + 1} = \frac{15 - 3}{4} = 3 \).
So, the coordinates of the point are \( (7, 3) \).
In simple words: The section formula helps us find a point that is three-quarters of the way along the line from the first point to the second point. Calculating this gives the coordinate values (7, 3).

Exam Tip: Verify your calculations by ensuring that the point \( (7, 3) \) lies on the straight line joining the two endpoints.

<3M>

Question 60. If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a parallelogram, taken in order, find the value of p.
Answer: In a parallelogram \( ABCD \), the diagonals bisect each other, which means the midpoint of diagonal \( AC \) is the same as the midpoint of diagonal \( BD \).
Midpoint of \( AC \):
\( \left( \frac{6 + 9}{2}, \frac{1 + 4}{2} \right) = \left( \frac{15}{2}, \frac{5}{2} \right) \)
Midpoint of \( BD \):
\( \left( \frac{8 + p}{2}, \frac{2 + 3}{2} \right) = \left( \frac{8 + p}{2}, \frac{5}{2} \right) \)
Since the midpoints are identical, we equate their x-coordinates:
\( \frac{15}{2} = \frac{8 + p}{2} \)

\( 15 = 8 + p \)

\( p = 7 \).
Thus, the value of \( p \) is 7.
In simple words: Since the diagonals of a parallelogram meet at their exact midpoints, the center coordinates calculated using both diagonal lines must be equal. Equating the horizontal values tells us that \( p \) must be 7.

Exam Tip: Avoid calculating the lengths of the sides to find variables in parallelograms. Using the midpoint of diagonals is much faster and less prone to calculation errors.

Question 61. If A(-5, 7), B(- 4, -5), C(-1, -6) and D(4, 5) are the vertices of a quadrilateral, find the area of the quadrilateral ABCD.
Answer: We can find the area of the quadrilateral \( ABCD \) by dividing it into two triangles, \( \triangle ABC \) and \( \triangle ACD \), by drawing diagonal \( AC \).
The total area is the sum of the areas of these two triangles.
Area of \( \triangle ABC \) with vertices \( A(-5, 7) \), \( B(-4, -5) \), and \( C(-1, -6) \):
\( \text{Area}_1 = \frac{1}{2} | -5(-5 - (-6)) + (-4)(-6 - 7) + (-1)(7 - (-5)) | \)
\( = \frac{1}{2} | -5(1) - 4(-13) - 1(12) | \)
\( = \frac{1}{2} | -5 + 52 - 12 | = \frac{1}{2} | 35 | = 17.5 \) sq. units.

Area of \( \triangle ACD \) with vertices \( A(-5, 7) \), \( C(-1, -6) \), and \( D(4, 5) \):
\( \text{Area}_2 = \frac{1}{2} | -5(-6 - 5) + (-1)(5 - 7) + 4(7 - (-6)) | \)
\( = \frac{1}{2} | -5(-11) - 1(-2) + 4(13) | \)
\( = \frac{1}{2} | 55 + 2 + 52 | = \frac{1}{2} | 109 | = 54.5 \) sq. units.

Total Area \( = \text{Area}_1 + \text{Area}_2 = 17.5 + 54.5 = 72 \) sq. units.
In simple words: To find the area of the quadrilateral, we draw a diagonal line to split it into two triangles. We calculate the area of both triangles using their corner positions and add them together, giving us 72 square units.

Exam Tip: Be consistent with the sign operations inside absolute value bars. Always simplify the terms inside the bracket before performing final division by 2.

Question 62. Find the value of k if the points A(2, 3), B(4, k) and C(6, -3) are collinear.
Answer: Since the points are collinear, the area of the triangle formed by them must be 0.
Using the triangle area formula:
\( \frac{1}{2} | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) | = 0 \)
Substitute the given points:
\( | 2(k - (-3)) + 4(-3 - 3) + 6(3 - k) | = 0 \)

\( | 2(k + 3) + 4(-6) + 18 - 6k | = 0 \)

\( | 2k + 6 - 24 + 18 - 6k | = 0 \)

\( | -4k | = 0 \implies k = 0 \).
So, the value of \( k \) is 0.
In simple words: Since the three points lie on the same straight line, they cannot form a triangle, meaning their area must be zero. Solving this equation shows that \( k \) is equal to 0.

Exam Tip: Collinear points can also be solved by equating slopes: \( \text{Slope of } AB = \text{Slope of } BC \). This method is often faster and avoids absolute values.

Question 63. Find the area of a triangle formed by the points A(5, 2), B(4, 7) and C (7, - 4).
Answer: Let the vertices of the triangle be \( (x_1, y_1) = (5, 2) \), \( (x_2, y_2) = (4, 7) \), and \( (x_3, y_3) = (7, -4) \).
Using the area formula:
Area \( = \frac{1}{2} | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) | \)

\( \text{Area} = \frac{1}{2} | 5(7 - (-4)) + 4(-4 - 2) + 7(2 - 7) | \)

\( \text{Area} = \frac{1}{2} | 5(11) + 4(-6) + 7(-5) | \)

\( \text{Area} = \frac{1}{2} | 55 - 24 - 35 | \)

\( \text{Area} = \frac{1}{2} | -4 | = 2 \) sq. units.
In simple words: Substituting the given coordinates into the triangle area formula shows that the space enclosed inside the triangle is exactly 2 square units.

Exam Tip: Ensure that you apply the absolute value at the end of the calculation to convert any negative results to a positive area value.

Question 64. In what ratio does the point (- 4, 6) divide the line segment joining the points A(- 6, 10) and B(3, - 8)?
Answer: Let the point \( P(-4, 6) \) divide the line segment joining \( A(-6, 10) \) and \( B(3, -8) \) internally in the ratio \( k : 1 \).
Using the section formula for the x-coordinate:
\( x_P = \frac{k x_2 + 1 x_1}{k + 1} \)

\( -4 = \frac{k(3) + 1(-6)}{k + 1} \)

\( -4(k + 1) = 3k - 6 \)

\( -4k - 4 = 3k - 6 \)

\( -7k = -2 \implies k = \frac{2}{7} \).
So, the ratio is \( 2 : 7 \).
In simple words: We assume the line is divided in a ratio of \( k : 1 \). Applying the coordinate division formula to the horizontal coordinates tells us that the ratio is 2 to 7.

Exam Tip: Verify your ratio by checking it with the y-coordinate using the same value of \( k \) to ensure both coordinates remain consistent.

<6M>

Question 65. If A(5, -1), B(-3, -2) and C(-1, 8) are the vertices of triangle ABC, find the length of median through A and the coordinates of the centroid.
Answer: Let us first find the coordinates of the midpoint of side \( BC \), which we call \( D \).
Midpoint \( D \) is calculated as:
\( D = \left( \frac{-3 + (-1)}{2}, \frac{-2 + 8}{2} \right) = \left( \frac{-4}{2}, \frac{6}{2} \right) = (-2, 3) \).
The median is the line segment connecting \( A(5, -1) \) and \( D(-2, 3) \). We find its length using the distance formula:
\( AD = \sqrt{(-2 - 5)^2 + (3 - (-1))^2} \)

\( AD = \sqrt{(-7)^2 + 4^2} = \sqrt{49 + 16} = \sqrt{65} \) units.
Now, we calculate the coordinates of the centroid of \( \triangle ABC \) using the centroid formula:
Centroid \( = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \)

\( \text{Centroid} = \left( \frac{5 + (-3) + (-1)}{3}, \frac{-1 + (-2) + 8}{3} \right) = \left( \frac{1}{3}, \frac{5}{3} \right) \).
So, the length of the median is \( \sqrt{65} \) units and the centroid is \( \left( \frac{1}{3}, \frac{5}{3} \right) \).
In simple words: The median starts from corner \( A \) and meets the midpoint of side \( BC \). We find the midpoint, measure its distance to \( A \) to get the median's length (\( \sqrt{65} \)), and then use the average corner positions to find the center (centroid) of the triangle as \( \left( \frac{1}{3}, \frac{5}{3} \right) \).

Exam Tip: Clearly label both parts of your answer (median length and centroid coordinates) to make it easy for the examiner to award full marks.

Question 66. Prove that the points (-3, 0), (1, -3) and (4,1) are the vertices of an isosceles right angled triangle. Find the area of this triangle.
Answer: Let the vertices be \( A(-3, 0) \), \( B(1, -3) \), and \( C(4, 1) \). We find the lengths of the three sides using the distance formula:
\( AB = \sqrt{(1 - (-3))^2 + (-3 - 0)^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \) units.
\( BC = \sqrt{(4 - 1)^2 + (1 - (-3))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \) units.
\( CA = \sqrt{(-3 - 4)^2 + (0 - 1)^2} = \sqrt{(-7)^2 + (-1)^2} = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2} \) units.
Since two sides are equal (\( AB = BC = 5 \)), \( \triangle ABC \) is an isosceles triangle.
Now, let us check Pythagoras' theorem:
\( AB^2 + BC^2 = 5^2 + 5^2 = 25 + 25 = 50 \)
And \( CA^2 = 50 \).
Since \( AB^2 + BC^2 = CA^2 \), the triangle is right-angled at \( B \).
Hence, \( \triangle ABC \) is an isosceles right-angled triangle.
The area of this right-angled triangle is:
Area \( = \frac{1}{2} \times \text{base} \times \text{height} \)

\( \text{Area} = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 5 \times 5 = 12.5 \) sq. units.
In simple words: We find that two side lengths are equal, proving the triangle is isosceles. Since the squares of these two sides add up to the square of the longest side, it is also a right-angled triangle. Its area is half of base times height, which is 12.5 square units.

Exam Tip: Clearly state both proofs (equal sides for isosceles, and the Pythagorean relation for the right angle) to secure full marks in long-answer questions.

Question 67. Prove that the co-ordinate of the centroid of the triangle whose vertices are \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) are \( \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \). Also deduce that the medians are concurrent.
Answer: Let \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \) be the vertices of the triangle \( ABC \).
Let \( D \) be the midpoint of side \( BC \). The coordinates of \( D \) are \( \left( \frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2} \right) \).
The centroid \( G \) is a point on the median \( AD \) that divides it internally in the ratio \( 2 : 1 \).
Using the internal section formula for point \( G \):
\( x_G = \frac{2 \left( \frac{x_2 + x_3}{2} \right) + 1 \cdot x_1}{2 + 1} = \frac{x_1 + x_2 + x_3}{3} \)
And,
\( y_G = \frac{2 \left( \frac{y_2 + y_3}{2} \right) + 1 \cdot y_1}{2 + 1} = \frac{y_1 + y_2 + y_3}{3} \).
So, the coordinates of \( G \) are \( \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \).
Similarly, if we find the coordinates of points dividing the medians through \( B \) and \( C \) in the ratio \( 2 : 1 \), we get the exact same coordinates. Since all three medians pass through this identical point \( G \), it proves that the medians of a triangle are concurrent (they meet at a single point).
In simple words: We find the midpoint of one side of the triangle and use the section formula in a 2 to 1 ratio on the median line to find the coordinates of the centroid. Repeating this for all three median lines yields the exact same center point, proving they all cross at the same spot.

Exam Tip: High-weightage proofs require clear step-by-step derivations. Drawing a labeled triangle with its medians greatly improves the presentation of your proof.

Question 68. Find the area of the triangle formed by joining the mid - points of the sides of the triangle whose vertices are (0, -1),(2, 1) and (0, 3). Find the ratio of the area of the triangle formed to the area of the given triangle.
Answer: Let the vertices of the given triangle be \( A(0, -1) \), \( B(2, 1) \), and \( C(0, 3) \).
First, we calculate the area of \( \triangle ABC \):
Area of \( \triangle ABC = \frac{1}{2} | 0(1 - 3) + 2(3 - (-1)) + 0(-1 - 1) | \)
\( = \frac{1}{2} | 0 + 2(4) + 0 | = \frac{1}{2} | 8 | = 4 \) sq. units.
Let \( D \), \( E \), and \( F \) be the midpoints of the sides \( BC \), \( CA \), and \( AB \) respectively.
Using the midpoint formula:
\( D = \left( \frac{2 + 0}{2}, \frac{1 + 3}{2} \right) = (1, 2) \)
\( E = \left( \frac{0 + 0}{2}, \frac{3 + (-1)}{2} \right) = (0, 1) \)
\( F = \left( \frac{0 + 2}{2}, \frac{-1 + 1}{2} \right) = (1, 0) \).
Now, we calculate the area of \( \triangle DEF \) formed by these midpoints:
Area of \( \triangle DEF = \frac{1}{2} | 1(1 - 0) + 0(0 - 2) + 1(2 - 1) | \)
\( = \frac{1}{2} | 1(1) + 0 + 1(1) | = \frac{1}{2} | 2 | = 1 \) sq. unit.
The ratio of the area of the midpoint triangle to the given triangle is:
\( \text{Ratio} = \frac{\text{Area of } \triangle DEF}{\text{Area of } \triangle ABC} = \frac{1}{4} = 1 : 4 \).
In simple words: The space inside the main triangle is 4 square units. The smaller triangle, made by joining the midpoints, has an area of 1 square unit. Comparing these two spaces gives us a ratio of 1 to 4.

Exam Tip: Remember that joining the midpoints of a triangle always divides it into 4 equal-area triangles. You can state this theorem directly to verify your calculated ratio of \( 1 : 4 \).

Question 69. Find the coordinates of the points of trisection of the line segment joining the points A(2, - 2) and B(- 7, 4).
Answer: Let the points of trisection be \( P \) and \( Q \). The point \( P \) divides the segment \( AB \) internally in the ratio \( 1 : 2 \), and the point \( Q \) divides it internally in the ratio \( 2 : 1 \).
Using the internal section formula for point \( P(1 : 2) \):
\( x_P = \frac{1(-7) + 2(2)}{1 + 2} = \frac{-7 + 4}{3} = -1 \)
\( y_P = \frac{1(4) + 2(-2)}{1 + 2} = \frac{4 - 4}{3} = 0 \)
So, the coordinates of \( P \) are \( (-1, 0) \).
Using the internal section formula for point \( Q(2 : 1) \):
\( x_Q = \frac{2(-7) + 1(2)}{2 + 1} = \frac{-14 + 2}{3} = -4 \)
\( y_Q = \frac{2(4) + 1(-2)}{2 + 1} = \frac{8 - 2}{3} = 2 \)
So, the coordinates of \( Q \) are \( (-4, 2) \).
Therefore, the points of trisection are \( (-1, 0) \) and \( (-4, 2) \).
In simple words: To cut the line into three equal pieces, we find two points. By using division ratios of 1 to 2 and 2 to 1 with the coordinates of the endpoints, we calculate these two points as (-1, 0) and (-4, 2).

Exam Tip: When calculating points of trisection, once you find the first point \( P \), the second point \( Q \) is simply the midpoint of the segment \( PB \). This shortcut can save time and reduce calculation steps.