OP Malhotra Class 11 Maths Solutions Chapter 15 Basic Concepts of Points and their Coordinates Exercise 15 (C)

Question 1. Find the area of the triangle whose vertices are
(i) (4, 2),(4, 5) and (-2, 2);
(ii) (0, 0),(-2, 3) and (10, 7);
(iii) (a, 0) (0, b) and (x, y).
Answer:
(i) To find the area of the triangle with vertices (4, 2), (4, 5), and (-2, 2), we use the shoelace formula. First, list the coordinates vertically, repeating the first point at the end:
\( \begin{matrix} 4 & 2 \\ 4 & 5 \\ -2 & 2 \\ 4 & 2 \end{matrix} \)
Now, multiply diagonally downwards and add these products: \( (4 \times 5) + (4 \times 2) + (-2 \times 2) = 20 + 8 - 4 = 24 \).
Next, multiply diagonally upwards and add these products: \( (2 \times 4) + (5 \times -2) + (2 \times 4) = 8 - 10 + 8 = 6 \).
The area is \( \frac { 1 }{ 2 } | (\text{sum of downward products}) - (\text{sum of upward products}) | \).
Required area \( = \frac { 1 }{ 2 } | (20 - 8) + (8 + 10) + (- 4 - 8) | \)
\( = \frac { 1 }{ 2 } | 12 + 18 - 12 | = \frac { 1 }{ 2 } | 18 | = 9 \) sq. units. The absolute value ensures the area is always positive.
(ii) For vertices (0, 0), (-2, 3), and (10, 7), we list the coordinates:
\( \begin{matrix} 0 & 0 \\ -2 & 3 \\ 10 & 7 \\ 0 & 0 \end{matrix} \)
Sum of downward products: \( (0 \times 3) + (-2 \times 7) + (10 \times 0) = 0 - 14 + 0 = -14 \).
Sum of upward products: \( (0 \times -2) + (3 \times 10) + (7 \times 0) = 0 + 30 + 0 = 30 \).
Required area \( = \frac { 1 }{ 2 } | (0 - 0) + (- 14 - 30) + (0 + 0) | \)
\( = \frac { 1 }{ 2 } | -14 - 30 | = \frac { 1 }{ 2 } | -44 | = 22 \) sq. units. This method is often called the shoelace formula.
(iii) For vertices (a, 0), (0, b), and (x, y), we list the coordinates:
\( \begin{matrix} a & 0 \\ 0 & b \\ x & y \\ a & 0 \end{matrix} \)
Sum of downward products: \( (a \times b) + (0 \times y) + (x \times 0) = ab + 0 + 0 = ab \).
Sum of upward products: \( (0 \times 0) + (b \times x) + (y \times a) = 0 + bx + ay \).
Required area \( = \frac { 1 }{ 2 } | (ab - 0) + (0 + bx) + (0 - ay) | \)
\( = \frac { 1 }{ 2 } | ab + bx - ay | \). To avoid a negative result in some cases, the expression is written as \( \frac { 1 }{ 2 } (bx + ay - ab) \) sq. units.
In simple words: To find the area of a triangle given its corners, you can use a formula that involves cross-multiplying the x and y coordinates. You write down the coordinates, repeat the first one, then multiply diagonally down and diagonally up, subtract the totals, and finally divide by two. Make sure the area is always a positive number.

🎯 Exam Tip: Remember the formula for the area of a triangle given vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) is \( \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \). The shoelace method is a visual way to apply this formula and can be less prone to sign errors if you follow the steps carefully.

Question 2. Find the area of the quadrilateral whose vertices are
(i) (1, 1),(7,- 3),(12, 2) and (7, 21);
(ii) (1, 1),(3, 4),(5, – 2) and (4, – 7).
Answer:
(i) To find the area of a quadrilateral with vertices (1, 1), (7, -3), (12, 2), and (7, 21), we use the shoelace formula, similar to triangles. List the coordinates vertically, repeating the first point at the end:
\( \begin{matrix} 1 & 1 \\ 7 & -3 \\ 12 & 2 \\ 7 & 21 \\ 1 & 1 \end{matrix} \)
Sum of downward products: \( (1 \times -3) + (7 \times 2) + (12 \times 21) + (7 \times 1) = -3 + 14 + 252 + 7 = 270 \).
Sum of upward products: \( (1 \times 7) + (-3 \times 12) + (2 \times 7) + (21 \times 1) = 7 - 36 + 14 + 21 = 6 \).
Required area \( = \frac { 1 }{ 2 } | (- 3 - 7) + (14 + 36) + (252 - 14) + (7 - 21) | \)
\( = \frac { 1 }{ 2 } | -10 + 50 + 238 - 14 | = \frac { 1 }{ 2 } | 264 | = 132 \) sq. units. This method efficiently calculates the area by considering the sequence of vertices.
(ii) For vertices (1, 1), (3, 4), (5, -2), and (4, -7), list the coordinates:
\( \begin{matrix} 1 & 1 \\ 3 & 4 \\ 5 & -2 \\ 4 & -7 \\ 1 & 1 \end{matrix} \)
Sum of downward products: \( (1 \times 4) + (3 \times -2) + (5 \times -7) + (4 \times 1) = 4 - 6 - 35 + 4 = -33 \).
Sum of upward products: \( (1 \times 3) + (4 \times 5) + (-2 \times 4) + (-7 \times 1) = 3 + 20 - 8 - 7 = 8 \).
Required area \( = \frac { 1 }{ 2 } | (4 - 3) + (- 6 - 20) + (- 35 + 8) + (4 + 7) | \)
\( = \frac { 1 }{ 2 } | 1 - 26 - 27 + 11 | = \frac { 1 }{ 2 } | -41 | = \frac { 41 }{ 2 } \) sq. units. It's important to list the vertices in order (clockwise or counter-clockwise) for this formula to work correctly.
In simple words: To find the area of a four-sided shape (quadrilateral), you use a similar cross-multiplication method as for a triangle. You write down the coordinates of the corners in order, repeat the first one, multiply diagonally, subtract the two totals, and divide by two. This is called the shoelace formula.

🎯 Exam Tip: When using the shoelace formula for a quadrilateral, ensure you list the vertices in consecutive order (either clockwise or counter-clockwise) to get the correct signed area. The absolute value then gives the true area.

Question 3. Show that the following points are collinear :
(i) (1, 4),(3, – 2) and (- 3, 16);
(ii) (- 5, 1), (5, 4) and (10, 7).
Answer:
(i) Points are collinear if the area of the triangle formed by them is zero. Let the given vertices be A(1, 4), B(3, -2), and C(-3, 16). We use the area formula:
\( \begin{matrix} 1 & 4 \\ 3 & -2 \\ -3 & 16 \\ 1 & 4 \end{matrix} \)
Sum of downward products: \( (1 \times -2) + (3 \times 16) + (-3 \times 4) = -2 + 48 - 12 = 34 \).
Sum of upward products: \( (4 \times 3) + (-2 \times -3) + (16 \times 1) = 12 + 6 + 16 = 34 \).
Area of \( \triangle ABC = \frac { 1 }{ 2 } | (- 2 - 12) + (48 - 6) + (- 12 - 16) | \)
\( = \frac { 1 }{ 2 } | -14 + 42 - 28 | = \frac { 1 }{ 2 } | 0 | = 0 \) sq. units.
Since the area of \( \triangle ABC \) is 0, the points A, B, and C lie on the same straight line.
\( \implies \) Therefore, the given points A, B, and C are collinear.
(ii) Let the given points be A(-5, 1), B(5, 4), and C(10, 7). We calculate the area of the triangle formed by these points:
\( \begin{matrix} -5 & 1 \\ 5 & 4 \\ 10 & 7 \\ -5 & 1 \end{matrix} \)
Sum of downward products: \( (-5 \times 4) + (5 \times 7) + (10 \times 1) = -20 + 35 + 10 = 25 \).
Sum of upward products: \( (1 \times 5) + (4 \times 10) + (7 \times -5) = 5 + 40 - 35 = 10 \).
Area of \( \triangle ABC = \frac { 1 }{ 2 } | (- 20 - 5) + (35 - 40) + (10 + 35) | \)
\( = \frac { 1 }{ 2 } | -25 - 5 + 45 | = \frac { 1 }{ 2 } | 15 | = \frac { 15 }{ 2 } \ne 0 \).
Since the area of \( \triangle ABC \) is not 0, the given points are not collinear. This calculation demonstrates that not all sets of points form a zero-area triangle.
In simple words: To check if three points are in a straight line (collinear), you can find the area of the triangle they would form. If the area is exactly zero, then the points are collinear, meaning they all lie on the same line. If the area is anything other than zero, they are not in a straight line.

🎯 Exam Tip: The condition for three points to be collinear is that the area of the triangle formed by them must be zero. Clearly state this condition and show the calculation of the area, concluding based on the result.

Question 4. If (7, a),(- 5, 2) and (3, 6) are collinear, find a.
Answer:Let the given collinear points be A(7, a), B(-5, 2), and C(3, 6). Since the points are collinear, the area of the triangle formed by them must be equal to 0. Area of \( \triangle ABC = 0 \). Using the shoelace formula for the area:
\( \begin{matrix} 7 & a \\ -5 & 2 \\ 3 & 6 \\ 7 & a \end{matrix} \)
Sum of downward products: \( (7 \times 2) + (-5 \times 6) + (3 \times a) = 14 - 30 + 3a = 3a - 16 \).
Sum of upward products: \( (a \times -5) + (2 \times 3) + (6 \times 7) = -5a + 6 + 42 = -5a + 48 \).
So, \( \frac { 1 }{ 2 } | (3a - 16) - (-5a + 48) | = 0 \).
\( \implies \frac { 1 }{ 2 } | 3a - 16 + 5a - 48 | = 0 \).
\( \implies \frac { 1 }{ 2 } | 8a - 64 | = 0 \).
\( \implies | 8a - 64 | = 0 \).
\( \implies 8a - 64 = 0 \).
\( \implies 8a = 64 \).
\( \implies a = 8 \). Thus, the value of 'a' that makes the points collinear is 8. The concept of collinearity is fundamental in coordinate geometry.
In simple words: When three points are in a straight line, the area of the triangle they form is zero. We use this fact to set up an equation with the unknown 'a'. By solving this equation, we can find the value of 'a' that makes the points collinear.

🎯 Exam Tip: When points are given as collinear, always set the area of the triangle formed by them to zero. Be careful with algebraic signs during calculation, especially when subtracting the sums of products.

Question 5. If the area of the quadrilateral whose angular points A, B, C, D taken in order are (1, 2), (- 5, 6), (7, – 4) and (-2, k) be zero, find the value of k.
Answer:Let the vertices of the quadrilateral be A(1, 2), B(-5, 6), C(7, -4), and D(-2, k). Since the area of the quadrilateral ABCD is zero, we can find the value of k using the shoelace formula. List the coordinates vertically, repeating the first point at the end:
\( \begin{matrix} 1 & 2 \\ -5 & 6 \\ 7 & -4 \\ -2 & k \\ 1 & 2 \end{matrix} \)
Sum of downward products: \( (1 \times 6) + (-5 \times -4) + (7 \times k) + (-2 \times 2) = 6 + 20 + 7k - 4 = 7k + 22 \).
Sum of upward products: \( (2 \times -5) + (6 \times 7) + (-4 \times -2) + (k \times 1) = -10 + 42 + 8 + k = k + 40 \).
Area of quadrilateral ABCD \( = \frac { 1 }{ 2 } | (7k + 22) - (k + 40) | \).
Given that the area is 0:
\( \frac { 1 }{ 2 } | (6 + 10) + (20 - 42) + (7k - 8) + (- 4 - k) | = 0 \)
\( \implies \frac { 1 }{ 2 } | 16 - 22 + 7k - 8 - 4 - k | = 0 \)
\( \implies \frac { 1 }{ 2 } | 6k - 18 | = 0 \).
\( \implies | 6k - 18 | = 0 \).
\( \implies 6k - 18 = 0 \).
\( \implies 6k = 18 \).
\( \implies k = 3 \). Thus, the value of k for which the area of the quadrilateral is zero is 3. When a quadrilateral has zero area, it implies its vertices are collinear, forming a degenerate quadrilateral.
In simple words: If a four-sided shape has zero area, it means all its corners lie on a single straight line. We use the area formula (the shoelace method) and set the area to zero. Then we solve the equation to find the missing coordinate 'k'.

🎯 Exam Tip: A quadrilateral with zero area implies that its vertices are collinear. Use the shoelace formula, ensure the vertices are in cyclic order, set the calculated area to zero, and then solve for the unknown variable.

Question 6. A, B, C are the points (-1, 5), (3, 1) and (5, 7) respectively. D, E, F, are the mid-points of BC, CA, AB respectively. Prove that △ABC = 4 △DEF.
Answer:Let A(-1, 5), B(3, 1), and C(5, 7) be the vertices of triangle ABC. First, we find the coordinates of the mid-points D, E, and F. D is the mid-point of BC: \( D = \left(\frac{3+5}{2}, \frac{1+7}{2}\right) = D(4, 4) \). E is the mid-point of AC: \( E = \left(\frac{-1+5}{2}, \frac{5+7}{2}\right) = E(2, 6) \). F is the mid-point of AB: \( F = \left(\frac{-1+3}{2}, \frac{5+1}{2}\right) = F(1, 3) \). Now, we calculate the area of \( \triangle ABC \):
\( \begin{matrix} -1 & 5 \\ 3 & 1 \\ 5 & 7 \\ -1 & 5 \end{matrix} \)
Sum of downward products: \( (-1 \times 1) + (3 \times 7) + (5 \times 5) = -1 + 21 + 25 = 45 \).
Sum of upward products: \( (5 \times 3) + (1 \times 5) + (7 \times -1) = 15 + 5 - 7 = 13 \).
Area of \( \triangle ABC = \frac { 1 }{ 2 } | 45 - 13 | = \frac { 1 }{ 2 } | 32 | = 16 \) sq. units. Next, we calculate the area of \( \triangle DEF \) with vertices D(4, 4), E(2, 6), and F(1, 3):
\( \begin{matrix} 4 & 4 \\ 2 & 6 \\ 1 & 3 \\ 4 & 4 \end{matrix} \)
Sum of downward products: \( (4 \times 6) + (2 \times 3) + (1 \times 4) = 24 + 6 + 4 = 34 \).
Sum of upward products: \( (4 \times 2) + (6 \times 1) + (3 \times 4) = 8 + 6 + 12 = 26 \).
Area of \( \triangle DEF = \frac { 1 }{ 2 } | 34 - 26 | = \frac { 1 }{ 2 } | 8 | = 4 \) sq. units. Comparing the areas: Area of \( \triangle ABC = 16 \) sq. units. Area of \( \triangle DEF = 4 \) sq. units. We can see that \( 16 = 4 \times 4 \), so Area of \( \triangle ABC = 4 \times \) Area of \( \triangle DEF \). This proves the relationship. This property is a known result relating a triangle to the triangle formed by its midpoints.
In simple words: We first found the middle points of each side of the big triangle ABC. This gave us a smaller triangle DEF. Then we calculated the area of both the big triangle and the small triangle using the coordinate formula. We found that the big triangle's area was four times larger than the small triangle's area, proving what the question asked.

🎯 Exam Tip: Remember the midpoint formula \( (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}) \) and apply the area of a triangle formula carefully. This problem demonstrates a key property: the triangle formed by joining the midpoints of the sides of a triangle has an area one-fourth of the original triangle.

Question 7. The straight lines y = m₁x + c₁, y = m2x + C2, and x = 0 intersect in the three points P, Q, and R. Find the area of the triangle PQR. What is the value of the area if c₁ = C₁?
Answer:Given equations of straight lines are: 1. \( y = m_1x + c_1 \) 2. \( y = m_2x + c_2 \) 3. \( x = 0 \) Point P is the intersection of equations (1) and (2). To find P, set \( m_1x + c_1 = m_2x + c_2 \). \( m_1x - m_2x = c_2 - c_1 \) \( x(m_1 - m_2) = c_2 - c_1 \)
\( \implies x = \frac{c_2 - c_1}{m_1 - m_2} \) Substitute x back into \( y = m_1x + c_1 \): \( y = m_1\left(\frac{c_2 - c_1}{m_1 - m_2}\right) + c_1 \) \( y = \frac{m_1c_2 - m_1c_1 + c_1(m_1 - m_2)}{m_1 - m_2} \) \( y = \frac{m_1c_2 - m_1c_1 + m_1c_1 - m_2c_1}{m_1 - m_2} \)
\( \implies y = \frac{m_1c_2 - m_2c_1}{m_1 - m_2} \) So, coordinates of P are \( \left(\frac{c_2 - c_1}{m_1 - m_2}, \frac{m_1c_2 - m_2c_1}{m_1 - m_2}\right) \). Point Q is the intersection of lines (2) and (3). Since \( x = 0 \), substitute into \( y = m_2x + c_2 \). \( y = m_2(0) + c_2 = c_2 \). So, coordinates of Q are \( (0, c_2) \). Point R is the intersection of lines (1) and (3). Since \( x = 0 \), substitute into \( y = m_1x + c_1 \). \( y = m_1(0) + c_1 = c_1 \). So, coordinates of R are \( (0, c_1) \). Now, we find the area of triangle PQR using the coordinates P\( \left(\frac{c_2 - c_1}{m_1 - m_2}, \frac{m_1c_2 - m_2c_1}{m_1 - m_2}\right) \), Q\( (0, c_2) \), and R\( (0, c_1) \). Let \( x_P = \frac{c_2 - c_1}{m_1 - m_2} \) and \( y_P = \frac{m_1c_2 - m_2c_1}{m_1 - m_2} \). The shoelace formula is \( \frac{1}{2} |x_P(y_Q - y_R) + x_Q(y_R - y_P) + x_R(y_P - y_Q)| \). Area of \( \triangle PQR = \frac{1}{2} \left| x_P(c_2 - c_1) + 0(c_1 - y_P) + 0(y_P - c_2) \right| \)
\( \implies \text{Area} = \frac{1}{2} \left| \left(\frac{c_2 - c_1}{m_1 - m_2}\right)(c_2 - c_1) \right| \)
\( \implies \text{Area} = \frac{1}{2} \frac{(c_2 - c_1)^2}{|m_1 - m_2|} \). We use the absolute value for \( m_1-m_2 \) to ensure the area is positive. If \( c_1 = c_2 \), then: Area \( = \frac{1}{2} \frac{(c_1 - c_1)^2}{|m_1 - m_2|} = \frac{1}{2} \frac{0^2}{|m_1 - m_2|} = 0 \). When \( c_1 = c_2 \), it means the two lines \( y = m_1x + c_1 \) and \( y = m_2x + c_1 \) share the same y-intercept. This implies that points P, Q, and R would be collinear, as Q and R would become the same point \( (0, c_1) \). The expression for the area is \( \frac{1}{2} \frac{(c_2 - c_1)^2}{|m_1 - m_2|} \). The coordinate calculation for the shoelace determinant for P, Q, R would be: \( \begin{matrix} \frac{c_2-c_1}{m_1-m_2} & \frac{m_1c_2-m_2c_1}{m_1-m_2} \\ 0 & c_2 \\ 0 & c_1 \\ \frac{c_2-c_1}{m_1-m_2} & \frac{m_1c_2-m_2c_1}{m_1-m_2} \end{matrix} \) Sum of downward products: \( \frac{c_2-c_1}{m_1-m_2} \cdot c_2 + 0 \cdot c_1 + 0 \cdot \frac{m_1c_2-m_2c_1}{m_1-m_2} = \frac{c_2(c_2-c_1)}{m_1-m_2} \) Sum of upward products: \( \frac{m_1c_2-m_2c_1}{m_1-m_2} \cdot 0 + c_2 \cdot 0 + c_1 \cdot \frac{c_2-c_1}{m_1-m_2} = \frac{c_1(c_2-c_1)}{m_1-m_2} \) Area \( = \frac{1}{2} \left| \frac{c_2(c_2-c_1)}{m_1-m_2} - \frac{c_1(c_2-c_1)}{m_1-m_2} \right| \) \( = \frac{1}{2} \left| \frac{(c_2-c_1)(c_2-c_1)}{m_1-m_2} \right| = \frac{1}{2} \frac{(c_2-c_1)^2}{|m_1-m_2|} \). If \( c_1 = c_2 \), the area becomes 0, meaning the three lines intersect at a single point (0, c1), or two of the "intersection" points are identical. When the y-intercepts are the same, the lines cross the y-axis at the same place.
In simple words: First, we find the exact points where each pair of lines crosses each other. These three points form the corners of our triangle PQR. Then, we use the special area formula for triangles to calculate the space inside PQR. If the 'c1' and 'c2' values are the same, it means all three lines meet at one single point on the y-axis, so the triangle formed would have no area at all.

🎯 Exam Tip: To find the area of a triangle formed by intersecting lines, first determine the coordinates of the three intersection points (vertices). Then apply the area of a triangle formula using these coordinates. Pay close attention to the conditions, like \( c_1 = c_2 \), which can simplify the area to zero, indicating collinear points.