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Math Notes For Class 12 Chapter 4 Ex 4.1 Solved & Quiz
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Math Notes For Class 12 Chapter 4 Ex 4.1 Solved & Quiz
Analytical Geometry (or Coordinate Geometry) bridges algebra and geometry by using a rectangular coordinate system. Exercise 4.1 focuses on position vectors of points in $\mathbb{R}^2$, distance measurements, division of line segments, properties of geometric figures, and loci of moving points.
| Concept / Quantity | Mathematical Formula | Key Conditions / Notes |
|---|---|---|
| Distance Formula | $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ | Distance between points $A(x_1, y_1)$ and $B(x_2, y_2)$ in a Cartesian plane. Always non-negative. |
| Midpoint Formula | $$M(x, y) = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$ | Special case of ratio formula where $k_1 : k_2 = 1 : 1$. |
| Internal Ratio Formula | $$P(x, y) = \left(\frac{k_1 x_2 + k_2 x_1}{k_1 + k_2}, \frac{k_1 y_2 + k_2 y_1}{k_1 + k_2}\right)$$ | Point $P$ lies between $A$ and $B$, dividing segment $AB$ in ratio $k_1 : k_2$. |
| External Ratio Formula | $$P(x, y) = \left(\frac{k_1 x_2 - k_2 x_1}{k_1 - k_2}, \frac{k_1 y_2 - k_2 y_1}{k_1 - k_2}\right)$$ | Point $P$ lies outside segment $AB$ along the line passing through them ($k_1 \neq k_2$). |
| Centroid of a Triangle | $$G(x, y) = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$$ | Point of intersection of the three medians of $\triangle ABC$. |
| Area of Triangular Region | $$\Delta = \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$$ | Vertices listed in counter-clockwise order yield a positive area. If area $= 0$, points are collinear. |
Question 1 (Part i): Find the distance between the points $A(3, 1)$ and $B(-2, -4)$, and determine the coordinates of the midpoint of segment $AB$.
Solution:
Let $A(x_1, y_1) = (3, 1)$ and $B(x_2, y_2) = (-2, -4)$.
Step 1: Calculate Distance $|AB|$
Using the distance formula:
$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
$$d = \sqrt{(-2 - 3)^2 + (-4 - 1)^2}$$
$$d = \sqrt{(-5)^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}$$
Step 2: Calculate Midpoint $M(x, y)$
Using the midpoint formula:
$$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$
$$M = \left(\frac{3 + (-2)}{2}, \frac{1 + (-4)}{2}\right) = \left(\frac{1}{2}, -\frac{3}{2}\right)$$
Answer: Distance $= 5\sqrt{2}$ units, Midpoint $= \left(\frac{1}{2}, -\frac{3}{2}\right)$
Question 1 (Part ii): Find the distance between the points $P(-\sqrt{5}, -\sqrt{3})$ and $Q(2\sqrt{5}, 3\sqrt{3})$, and determine the midpoint of line segment $PQ$.
Solution:
Let $P(x_1, y_1) = (-\sqrt{5}, -\sqrt{3})$ and $Q(x_2, y_2) = (2\sqrt{5}, 3\sqrt{3})$.
Step 1: Calculate Distance $|PQ|$
$$d = \sqrt{[2\sqrt{5} - (-\sqrt{5})]^2 + [3\sqrt{3} - (-\sqrt{3})]^2}$$
$$d = \sqrt{(2\sqrt{5} + \sqrt{5})^2 + (3\sqrt{3} + \sqrt{3})^2}$$
$$d = \sqrt{(3\sqrt{5})^2 + (4\sqrt{3})^2}$$
$$d = \sqrt{(9 \cdot 5) + (16 \cdot 3)} = \sqrt{45 + 48} = \sqrt{93}$$
Step 2: Calculate Midpoint $M(x, y)$
$$M = \left(\frac{-\sqrt{5} + 2\sqrt{5}}{2}, \frac{-\sqrt{3} + 3\sqrt{3}}{2}\right) = \left(\frac{\sqrt{5}}{2}, \frac{2\sqrt{3}}{2}\right) = \left(\frac{\sqrt{5}}{2}, \sqrt{3}\right)$$
Answer: Distance $= \sqrt{93}$ units, Midpoint $= \left(\frac{\sqrt{5}}{2}, \sqrt{3}\right)$
Question 2 (Part i): Find the point $P(x,y)$ that divides the line segment joining $A(2, 3)$ and $B(7, 8)$ internally in the ratio $3:2$.
Solution:
Here, $A(x_1, y_1) = (2, 3)$, $B(x_2, y_2) = (7, 8)$, and ratio $k_1 : k_2 = 3 : 2$.
Applying the internal ratio formula:
$$x = \frac{k_1 x_2 + k_2 x_1}{k_1 + k_2} = \frac{3(7) + 2(2)}{3 + 2} = \frac{21 + 4}{5} = \frac{25}{5} = 5$$
$$y = \frac{k_1 y_2 + k_2 y_1}{k_1 + k_2} = \frac{3(8) + 2(3)}{3 + 2} = \frac{24 + 6}{5} = \frac{30}{5} = 6$$
Answer: Coordinates of point $P = (5, 6)$
Question 2 (Part ii): Find the point $P(x,y)$ that divides the line segment joining $A(5, -2)$ and $B(-1, 4)$ externally in the ratio $4:1$.
Solution:
Here, $A(x_1, y_1) = (5, -2)$, $B(x_2, y_2) = (-1, 4)$, and ratio $k_1 : k_2 = 4 : 1$.
Applying the external ratio formula:
$$x = \frac{k_1 x_2 - k_2 x_1}{k_1 - k_2} = \frac{4(-1) - 1(5)}{4 - 1} = \frac{-4 - 5}{3} = \frac{-9}{3} = -3$$
$$y = \frac{k_1 y_2 - k_2 y_1}{k_1 - k_2} = \frac{4(4) - 1(-2)}{4 - 1} = \frac{16 + 2}{3} = \frac{18}{3} = 6$$
Answer: Coordinates of point $P = (-3, 6)$
Question 3 (Part i): Show that the points $A(1, 1)$, $B(4, 5)$, and $C(9, -5)$ form a right-angled triangle.
Solution:
Calculate the square of the lengths of all three sides of $\triangle ABC$:
$$|AB|^2 = (4 - 1)^2 + (5 - 1)^2 = (3)^2 + (4)^2 = 9 + 16 = 25$$
$$|BC|^2 = (9 - 4)^2 + (-5 - 5)^2 = (5)^2 + (-10)^2 = 25 + 100 = 125$$
$$|AC|^2 = (9 - 1)^2 + (-5 - 1)^2 = (8)^2 + (-6)^2 = 64 + 36 = 100$$
Observe the relationship between side lengths:
$$|AB|^2 + |AC|^2 = 25 + 100 = 125 = |BC|^2$$
Since the sum of squares of two sides equals the square of the third side, by the converse of the Pythagorean Theorem, $\triangle ABC$ is a right-angled triangle with the right angle at vertex $A$.
Answer: $$\text{Proved: } |AB|^2 + |AC|^2 = |BC|^2 \implies \triangle ABC \text{ is right-angled.}$$
Question 3 (Part ii): Determine whether the points $P(-1, -1)$, $Q(2, 3)$, and $R(8, 11)$ are collinear.
Solution:
Method (Distance Summation):
Compute distances between all pairs of points:
$$|PQ| = \sqrt{[2 - (-1)]^2 + [3 - (-1)]^2} = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$$
$$|QR| = \sqrt{(8 - 2)^2 + (11 - 3)^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$
$$|PR| = \sqrt{[8 - (-1)]^2 + [11 - (-1)]^2} = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15$$
Now check if $|PQ| + |QR| = |PR|$:
$$5 + 10 = 15$$
Since the sum of the distances between two pairs equals the distance between the endpoints, the points lie on the same straight line.
Answer: $$\text{Points } P, Q, \text{ and } R \text{ are collinear.}$$
Question 4 (Part i): Find the centroid of the triangle with vertices $A(-2, 3)$, $B(4, -1)$, and $C(1, 7)$.
Solution:
Given $A(x_1, y_1) = (-2, 3)$, $B(x_2, y_2) = (4, -1)$, and $C(x_3, y_3) = (1, 7)$.
Using the Centroid Formula:
$$G(x, y) = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$$
$$x = \frac{-2 + 4 + 1}{3} = \frac{3}{3} = 1$$
$$y = \frac{3 + (-1) + 7}{3} = \frac{9}{3} = 3$$
Answer: Centroid $G = (1, 3)$
Question 4 (Part ii): Calculate the exact area of the triangle formed by the vertices $A(0, 0)$, $B(6, 2)$, and $C(3, 8)$.
Solution:
Using the area formula for a triangle in matrix form:
$$\text{Area } \Delta = \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$$
$$\Delta = \frac{1}{2} \begin{vmatrix} 0 & 0 & 1 \\ 6 & 2 & 1 \\ 3 & 8 & 1 \end{vmatrix}$$
Expanding along Row 1:
$$\Delta = \frac{1}{2} \left[ 0 - 0 + 1 \cdot (6 \cdot 8 - 3 \cdot 2) \right]$$
$$\Delta = \frac{1}{2} [1 \cdot (48 - 6)] = \frac{1}{2} (42) = 21$$
Answer: Area $= 21\text{ square units}$
Question 5 (Part i): Three vertices of a parallelogram taken in order are $A(-1, 2)$, $B(3, 5)$, and $C(7, 1)$. Find the coordinates of the fourth vertex $D(x, y)$.
Solution:
In a parallelogram $ABCD$, the diagonals $AC$ and $BD$ bisect each other, meaning they share the exact same midpoint $M$.
Step 1: Find Midpoint of $AC$
$$M_{AC} = \left(\frac{-1 + 7}{2}, \frac{2 + 1}{2}\right) = \left(3, \frac{3}{2}\right)$$