GSEB Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.3

Question 1. Find the area of the triangle whose vertices are:
(i) (2, 3), (-1, 0), (2, -4)
(ii) (-5, -1), (3, -5), (5, 2)
Answer:
(i) Suppose the specified points are A(2, 3), B(-1, 0), and C(2, -4).
Here, we have these values:
\( x_1 = 2, y_1 = 3 \)
\( x_2 = -1, y_2 = 0 \)
\( x_3 = 2, y_3 = -4 \)
Next, the area of \( \Delta ABC \) is determined by the formula:
Area of \( \Delta ABC = \frac {1}{2} [x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)] \)
\( = \frac {1}{2} [2(0 - (-4)) + (-1)(-4 - 3) + 2(3 - 0)] \)
\( = \frac {1}{2} [2(0 + 4) + (-1)(-7) + 2(3)] \)
\( = \frac {1}{2} [8 + 7 + 6] \)
\( = \frac {1}{2} \times 21 \)
\( = \frac {21}{2} \) sq. units
(ii) Let the specified points be A(-5, -1), B(3, -5), and C(5, 2).
Here, we have these values:
\( x_1 = -5, y_1 = -1 \)
\( x_2 = 3, y_2 = -5 \)
\( x_3 = 5, y_3 = 2 \)
Next, the area of \( \Delta ABC \) is calculated:
Area of \( \Delta ABC = \frac {1}{2} [x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)] \)
\( = \frac {1}{2} [-5(-5 - 2) + 3(2 - (-1)) + 5(-1 - (-5))] \)
\( = \frac {1}{2} [-5(-7) + 3(2 + 1) + 5(-1 + 5)] \)
\( = \frac {1}{2} [35 + 9 + 20] \)
\( = \frac {1}{2} \times 64 \)
\( = 32 \) sq. units
In simple words: To find a triangle's area, you use a special formula with the coordinates of its corners. You substitute the x and y values into the formula and then solve it. The area must always be a positive number.

Exam Tip: Remember the area formula for a triangle with given vertices, and be careful with negative signs during calculation, especially when subtracting y-coordinates.

Question 2. In each of the following find the value of 'k', for which the points are collinear.
(i) (7, -2), (5, 1), (3, k)
(ii) (8, 1), (k, -4), (2, -5)
Answer:
(i) Let the specified points be A(7, -2), B(5, 1), and C(3, k).
Here, we have these values:
\( x_1 = 7, y_1 = -2 \)
\( x_2 = 5, y_2 = 1 \)
\( x_3 = 3, y_3 = k \)
Next, the area of \( \Delta ABC \) is calculated:
Area of \( \Delta ABC = \frac {1}{2} [x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)] \)
\( = \frac {1}{2} [7(1 - k) + 5(k - (-2)) + 3(-2 - 1)] \)
\( = \frac {1}{2} [7 - 7k + 5(k + 2) + 3(-3)] \)
\( = \frac {1}{2} [7 - 7k + 5k + 10 - 9] \)
\( = \frac {1}{2} [8 - 2k] \)
\( = 4 - k \)
If the points are collinear, it means they lie on a straight line, so the area of the triangle formed by them must be 0.
Therefore, we set the area to zero:
\( 4 - k = 0 \)
\( \implies k = 4 \)
(ii) Let the specified points be A(8, 1), B(k, -4), and C(2, -5).
Here, we have these values:
\( x_1 = 8, y_1 = 1 \)
\( x_2 = k, y_2 = -4 \)
\( x_3 = 2, y_3 = -5 \)
Next, the area of \( \Delta ABC \) is calculated:
Area of \( \Delta ABC = \frac {1}{2} [x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)] \)
\( = \frac {1}{2} [8(-4 - (-5)) + k(-5 - 1) + 2(1 - (-4))] \)
\( = \frac {1}{2} [8(-4 + 5) + k(-6) + 2(1 + 4)] \)
\( = \frac {1}{2} [8(1) - 6k + 2(5)] \)
\( = \frac {1}{2} [8 - 6k + 10] \)
\( = \frac {1}{2} [18 - 6k] \)
\( = 9 - 3k \)
If the points are collinear, it means they lie on a straight line, so the area of the triangle formed by them must be 0.
Therefore, we set the area to zero:
\( 9 - 3k = 0 \)
\( \implies 3k = 9 \)
\( \implies k = 3 \)
In simple words: When three points are in a straight line (collinear), the triangle they form has zero area. So, we use the area formula, set it equal to zero, and then solve for 'k'.

Exam Tip: Remember that for collinear points, the area of the triangle formed by them is always zero. This is a key condition for solving such problems.

Question 3. 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 area of the given triangle.
Answer: Let A(0, -1), B(2, 1), and C(0, 3) be the vertices of \( \Delta ABC \). Let D, E, F be the mid-points of sides BC, CA, and AB respectively.
Then, the coordinates of D, E, and F are (1, 2), (0, 1), and (1, 0) respectively.
Next, calculate the Area of \( \Delta ABC \):
Area of \( \Delta ABC = \frac {1}{2} [x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)] \)
\( = \frac {1}{2} [0(1 - 3) + 2(3 - (-1)) + 0(-1 - 1)] \)
\( = \frac {1}{2} [0(-2) + 2(3 + 1) + 0(-2)] \)
\( = \frac {1}{2} [0 + 8 + 0] \)
\( = \frac {1}{2} \times 8 \)
\( = 4 \) sq. units
Now, calculate the Area of \( \Delta DEF \):
Area of \( \Delta DEF = \frac {1}{2} [x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)] \)
\( = \frac {1}{2} [1(1 - 0) + 0(0 - 2) + 1(2 - 1)] \)
\( = \frac {1}{2} [1(1) + 0(-2) + 1(1)] \)
\( = \frac {1}{2} [1 + 0 + 1] \)
\( = \frac {1}{2} [2] \)
\( = 1 \) sq. unit
Therefore, the ratio of Area of \( \Delta DEF \) to Area of \( \Delta ABC \) is \( 1 : 4 \).
In simple words: First, find the middle points of each side of the main triangle. Then, calculate the area of both the main triangle and the smaller triangle formed by these middle points. Finally, compare their areas to find the ratio.

Exam Tip: Remember the midpoint formula: \( \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \). The area of the triangle formed by joining the midpoints of the sides of a given triangle is always one-fourth the area of the given triangle.

Question 4. Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3). [National Olympiad]
Answer: Let the vertices of the quadrilateral be A(-4, -2), B(-3, -5), C(3, -2), and D(2, 3).
We can find the area of the quadrilateral by dividing it into two triangles, \( \Delta ABC \) and \( \Delta ACD \), and then summing their areas.
First, calculate the Area of \( \Delta ABC \):
Here, we have:
\( x_1 = -4, y_1 = -2 \)
\( x_2 = -3, y_2 = -5 \)
\( x_3 = 3, y_3 = -2 \)
Area of \( \Delta ABC = \frac {1}{2} [x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)] \)
\( = \frac {1}{2} [-4(-5 - (-2)) + (-3)(-2 - (-2)) + 3(-2 - (-5))] \)
\( = \frac {1}{2} [-4(-5 + 2) + (-3)(-2 + 2) + 3(-2 + 5)] \)
\( = \frac {1}{2} [-4 \times -3 + (-3) \times 0 + 3 \times 3] \)
\( = \frac {1}{2} [12 + 0 + 9] \)
\( = \frac {1}{2} \times 21 \)
\( = \frac {21}{2} \) sq. units
Next, calculate the Area of \( \Delta ACD \):
Here, we have:
\( x_1 = -4, y_1 = -2 \)
\( x_2 = 3, y_2 = -2 \)
\( x_3 = 2, y_3 = 3 \)
Area of \( \Delta ACD = \frac {1}{2} [x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)] \)
\( = \frac {1}{2} [-4(-2 - 3) + 3(3 - (-2)) + 2(-2 - (-2))] \)
\( = \frac {1}{2} [-4(-5) + 3(3 + 2) + 2(-2 + 2)] \)
\( = \frac {1}{2} [20 + 3(5) + 2(0)] \)
\( = \frac {1}{2} [20 + 15 + 0] \)
\( = \frac {1}{2} \times 35 \)
\( = \frac {35}{2} \) sq. units
Hence, the total area of quadrilateral ABCD is the sum of the areas of these two triangles:
Area of quadrilateral ABCD \( = \text{Area}(\Delta ABC) + \text{Area}(\Delta ACD) \)
\( = \frac {21}{2} + \frac {35}{2} \)
\( = \frac {1}{2}(21 + 35) \)
\( = \frac {1}{2} \times 56 \)
\( = 28 \) sq. units
In simple words: To find the area of a four-sided shape (quadrilateral), you can split it into two triangles. Calculate the area of each triangle using the coordinate formula, and then add those two areas together to get the total area of the quadrilateral.

Exam Tip: Always make sure to take the vertices in order when calculating the area of a quadrilateral by dividing it into triangles. Using the incorrect order might result in an incorrect sign for the area of one of the triangles, leading to an error in the final sum.

Question 5. You have studied in Class IX, that a median of a triangle divides it into two triangles of equal areas. Verify this result for \( \Delta ABC \) whose vertices are A(4, -6), B(3, -2) and C(5, 2).
Answer: According to the question, AD is the median of \( \Delta ABC \), so D is the mid-point of BC.
Therefore, the coordinates of D are \( \left(\frac{3+5}{2}, \frac{-2+2}{2}\right) \), which simplifies to (4, 0).
Now, we calculate the Area of \( \Delta ABD \):
Let the points be A(4, -6), B(3, -2), and D(4, 0).
\( x_1 = 4, y_1 = -6 \)
\( x_2 = 3, y_2 = -2 \)
\( x_3 = 4, y_3 = 0 \)
Area of \( \Delta ABD = \frac {1}{2} [x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)] \)
\( = \frac {1}{2} [4(-2 - 0) + 3(0 - (-6)) + 4(-6 - (-2))] \)
\( = \frac {1}{2} [4(-2) + 3(6) + 4(-4)] \)
\( = \frac {1}{2} [-8 + 18 - 16] \)
\( = \frac {1}{2} [-6] \)
Since the area of a triangle must be positive, we take the absolute value:
Area of \( \Delta ABD = 3 \) sq. units
Next, calculate the Area of \( \Delta ADC \):
Let the points be A(4, -6), D(4, 0), and C(5, 2).
\( x_1 = 4, y_1 = -6 \)
\( x_2 = 4, y_2 = 0 \)
\( x_3 = 5, y_3 = 2 \)
Area of \( \Delta ADC = \frac {1}{2} [x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)] \)
\( = \frac {1}{2} [4(0 - 2) + 4(2 - (-6)) + 5(-6 - 0)] \)
\( = \frac {1}{2} [4(-2) + 4(8) + 5(-6)] \)
\( = \frac {1}{2} [-8 + 32 - 30] \)
\( = \frac {1}{2} [-6] \)
Since the area of a triangle must be positive, we take the absolute value:
Area of \( \Delta ADC = 3 \) sq. units
Hence, since Area of \( \Delta ABD = 3 \) sq. units and Area of \( \Delta ADC = 3 \) sq. units, we can confirm that the median of the triangle divides it into two triangles of equal areas.
In simple words: First, find the midpoint of one side to locate where the median touches. Then, calculate the area of the two smaller triangles formed by this median. If their areas are the same, the result is verified.

Exam Tip: Remember that a median connects a vertex to the midpoint of the opposite side. When verifying properties like equal areas, ensure your midpoint calculation is correct before proceeding with area calculations.