OP Malhotra Class 11 Maths Solutions Chapter 15 Basic Concepts of Points and their Coordinates Test

Question 1. Show that the points (4, 4), (3, 5) and (- 1, – 1) are the vertices of a right triangle.
Answer: Let the three given points be A(4, 4), B(3, 5), and C(-1, -1). We need to calculate the distance between each pair of points using the distance formula \( d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \).
First, find the distance AB:
\( AB = \sqrt{(3-4)^2 + (5-4)^2} \)
\( AB = \sqrt{(-1)^2 + (1)^2} \)
\( AB = \sqrt{1 + 1} \)
\( AB = \sqrt{2} \)
Next, find the distance BC:
\( BC = \sqrt{(-1-3)^2 + (-1-5)^2} \)
\( BC = \sqrt{(-4)^2 + (-6)^2} \)
\( BC = \sqrt{16 + 36} \)
\( BC = \sqrt{52} \)
\( BC = 2\sqrt{13} \)
Then, find the distance CA:
\( CA = \sqrt{(-1-4)^2 + (-1-4)^2} \)
\( CA = \sqrt{(-5)^2 + (-5)^2} \)
\( CA = \sqrt{25 + 25} \)
\( CA = \sqrt{50} \)
\( CA = 5\sqrt{2} \)
Now, we check if these lengths satisfy the Pythagorean theorem \( a^2 + b^2 = c^2 \). We square each length:
\( AB^2 = (\sqrt{2})^2 = 2 \)
\( BC^2 = (\sqrt{52})^2 = 52 \)
\( CA^2 = (\sqrt{50})^2 = 50 \)
We can see that \( AB^2 + CA^2 = 2 + 50 = 52 \). This is equal to \( BC^2 \).
Since \( AB^2 + CA^2 = BC^2 \), the points A, B, and C form a right-angled triangle where the right angle is at point A, opposite the hypotenuse BC.
In simple words: We find the length of each side of the triangle. If the square of the longest side equals the sum of the squares of the other two sides, then it's a right-angled triangle. This is the Pythagorean theorem in action.

🎯 Exam Tip: To show points form a right triangle, calculate the square of the distance between all three pairs of points. Then check if the sum of two smaller squares equals the largest square. Remember to label your points clearly for easy understanding.

Question 2. A quadrilateral has the vertices at the points (-4, 2), (2, 6), (8, 5) and (9, – 7). Show that the mid-points of the sides of this quadrilateral are the vertices of a parallelogram.
Answer: Let the vertices of the quadrilateral be A(-4, 2), B(2, 6), C(8, 5), and D(9, -7). To show that the mid-points of its sides form a parallelogram, we first find the coordinates of these mid-points.
Let E, F, G, and H be the mid-points of sides AB, BC, CD, and AD, respectively. We use the mid-point formula \( \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) \):
Coordinates of E (mid-point of AB):
\( E = \left( \frac{-4+2}{2}, \frac{2+6}{2} \right) = \left( \frac{-2}{2}, \frac{8}{2} \right) = (-1, 4) \)
Coordinates of F (mid-point of BC):
\( F = \left( \frac{2+8}{2}, \frac{6+5}{2} \right) = \left( \frac{10}{2}, \frac{11}{2} \right) = \left( 5, \frac{11}{2} \right) \)
Coordinates of G (mid-point of CD):
\( G = \left( \frac{8+9}{2}, \frac{5+(-7)}{2} \right) = \left( \frac{17}{2}, \frac{-2}{2} \right) = \left( \frac{17}{2}, -1 \right) \)
Coordinates of H (mid-point of AD):
\( H = \left( \frac{9+(-4)}{2}, \frac{-7+2}{2} \right) = \left( \frac{5}{2}, \frac{-5}{2} \right) \)
Now we have the vertices of the new quadrilateral EFGH: E(-1, 4), F(5, 11/2), G(17/2, -1), H(5/2, -5/2).
A quadrilateral is a parallelogram if its opposite sides are equal in length, or if its diagonals bisect each other (meaning they have the same mid-point). Let's check the mid-points of the diagonals EG and FH.
Mid-point of diagonal EG:
\( M_{EG} = \left( \frac{-1+\frac{17}{2}}{2}, \frac{4+(-1)}{2} \right) \)
\( M_{EG} = \left( \frac{-\frac{2}{2}+\frac{17}{2}}{2}, \frac{3}{2} \right) \)
\( M_{EG} = \left( \frac{\frac{15}{2}}{2}, \frac{3}{2} \right) = \left( \frac{15}{4}, \frac{3}{2} \right) \)
Mid-point of diagonal FH:
\( M_{FH} = \left( \frac{5+\frac{5}{2}}{2}, \frac{\frac{11}{2}+\left(-\frac{5}{2}\right)}{2} \right) \)
\( M_{FH} = \left( \frac{\frac{10}{2}+\frac{5}{2}}{2}, \frac{\frac{6}{2}}{2} \right) \)
\( M_{FH} = \left( \frac{\frac{15}{2}}{2}, \frac{3}{2} \right) = \left( \frac{15}{4}, \frac{3}{2} \right) \)
Since the mid-point of diagonal EG and the mid-point of diagonal FH are the same, the diagonals bisect each other. This is a key property of a parallelogram.
Therefore, the quadrilateral EFGH formed by the mid-points is a parallelogram.
In simple words: First, find the exact middle point of each side of the main shape. These four middle points create a new shape. If the diagonals (lines from corner to opposite corner) of this new shape cross exactly in the middle of each other, then the new shape is a parallelogram.

🎯 Exam Tip: To prove a quadrilateral is a parallelogram using coordinates, the easiest method is often to show that the mid-points of its diagonals are identical. This implies the diagonals bisect each other, which is a sufficient condition for a parallelogram.

Question 3. Show that the points (a, b + c),(b, c + a),(c, a + b) are collinear.
Answer: For three points to be collinear (lie on the same straight line), the area of the triangle formed by these three points must be zero. Let the three points be A(a, b+c), B(b, c+a), and C(c, a+b).
The area of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is given by the formula:
\( \text{Area} = \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)| \)
Substitute the coordinates of A, B, and C into this formula:
\( \text{Area} = \frac{1}{2} |a((c+a)-(a+b)) + b((a+b)-(b+c)) + c((b+c)-(c+a))| \)
\( \text{Area} = \frac{1}{2} |a(c+a-a-b) + b(a+b-b-c) + c(b+c-c-a)| \)
\( \text{Area} = \frac{1}{2} |a(c-b) + b(a-c) + c(b-a)| \)
Now, expand the terms:
\( \text{Area} = \frac{1}{2} |ac - ab + ba - bc + cb - ca| \)
Rearrange and combine like terms:
\( \text{Area} = \frac{1}{2} |ac - ca - ab + ba - bc + cb| \)
\( \text{Area} = \frac{1}{2} |0 + 0 + 0| \)
\( \text{Area} = \frac{1}{2} \times 0 \)
\( \text{Area} = 0 \)
Since the area of the triangle formed by points A, B, and C is 0, the three points are collinear. This method is a reliable way to check for collinearity in coordinate geometry.
In simple words: Imagine drawing a triangle using these three points. If the points all fall on a single straight line, you can't actually make a "real" triangle, so its area will be zero. We use a math formula to show that the area is indeed zero, which proves they are on the same line.

🎯 Exam Tip: When asked to show points are collinear, the most common and robust method is to calculate the area of the triangle formed by them. If the area is zero, the points are collinear. Be careful with algebraic expansion and signs.

Question 4. Centroid of a triangle is (1, 4) and two of its vertices are (4, – 3) and (- 9, 7). Find the area of the triangle.
Answer: Let the two given vertices be \( A(4, -3) \) and \( B(-9, 7) \). Let the third vertex be \( C(\alpha, \beta) \). The centroid of the triangle, G, is given as \( (1, 4) \).
The formula for the centroid \( (x_G, y_G) \) of a triangle with vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) is:
\( (x_G, y_G) = \left( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3} \right) \)
Substitute the known values:
\( (1, 4) = \left( \frac{4 + (-9) + \alpha}{3}, \frac{-3 + 7 + \beta}{3} \right) \)
\( (1, 4) = \left( \frac{\alpha - 5}{3}, \frac{\beta + 4}{3} \right) \)
Now, we equate the x-coordinates and y-coordinates to find \( \alpha \) and \( \beta \):
For x-coordinate:
\( 1 = \frac{\alpha - 5}{3} \)
\( 3 = \alpha - 5 \)
\( \implies \alpha = 3 + 5 \)
\( \implies \alpha = 8 \)
For y-coordinate:
\( 4 = \frac{\beta + 4}{3} \)
\( 12 = \beta + 4 \)
\( \implies \beta = 12 - 4 \)
\( \implies \beta = 8 \)
So, the coordinates of the third vertex C are \( (8, 8) \).
Now that we have all three vertices A(4, -3), B(-9, 7), and C(8, 8), we can find the area of the triangle using the area formula:
\( \text{Area} = \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)| \)
\( \text{Area} = \frac{1}{2} |4(7-8) + (-9)(8-(-3)) + 8(-3-7)| \)
\( \text{Area} = \frac{1}{2} |4(-1) + (-9)(11) + 8(-10)| \)
\( \text{Area} = \frac{1}{2} |-4 - 99 - 80| \)
\( \text{Area} = \frac{1}{2} |-183| \)
\( \text{Area} = \frac{183}{2} \)
\( \text{Area} = 91.5 \) square units.
In simple words: A triangle's centroid is like its balancing point. If you know two corners and the balancing point, you can work backward to find the third corner. Once you have all three corners, you use a special formula to calculate the size of the triangle's surface.

🎯 Exam Tip: When the centroid and two vertices are given, always find the third vertex first using the centroid formula. After that, use the standard triangle area formula with all three vertices. Remember to take the absolute value for the area, as area cannot be negative.

Question 5. Find the third vertex of a triangle if two of its vertices are (- 1, 4) and (5, 2) and the medians through these vertices meet at (0, – 3).
Answer: The point where the medians of a triangle meet is called the centroid. So, the centroid G is given as (0, -3). Let the two known vertices be A(-1, 4) and B(5, 2). Let the third vertex be C\( (\alpha, \beta) \).
The formula for the centroid \( (x_G, y_G) \) of a triangle with vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) is:
\( (x_G, y_G) = \left( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3} \right) \)
Substitute the known values:
\( (0, -3) = \left( \frac{-1 + 5 + \alpha}{3}, \frac{4 + 2 + \beta}{3} \right) \)
\( (0, -3) = \left( \frac{4 + \alpha}{3}, \frac{6 + \beta}{3} \right) \)
Now, we equate the x-coordinates and y-coordinates to solve for \( \alpha \) and \( \beta \):
For x-coordinate:
\( 0 = \frac{4 + \alpha}{3} \)
\( 0 = 4 + \alpha \)
\( \implies \alpha = -4 \)
For y-coordinate:
\( -3 = \frac{6 + \beta}{3} \)
\( -9 = 6 + \beta \)
\( \implies \beta = -9 - 6 \)
\( \implies \beta = -15 \)
Therefore, the coordinates of the third vertex C are \( (-4, -15) \). This vertex helps complete the triangle and define its centroid.
In simple words: The problem gives you two corners of a triangle and its "balance point" (the centroid). Using a special formula that connects all three corners to the balance point, you can work out where the missing third corner is located.

🎯 Exam Tip: Remember that the intersection point of the medians is the centroid. This question directly tests your understanding and application of the centroid formula to find a missing vertex. Double-check your algebraic steps, especially with negative numbers.