CBSE Class 11 Mathematics Chapter 10: Straight Lines NCERT Solutions

Let the vertices of the quadrilateral be A(-4, 5), B(0, 7), C(5, -5), and D(-4, -2). We can plot these points on the Cartesian plane and connect them to form the quadrilateral ABCD.

To find the area of the quadrilateral ABCD, we can divide it into two triangles by drawing a diagonal, for example, AC. The area of the quadrilateral will be the sum of the areas of triangle ABC and triangle ACD.

The formula for the area of a triangle with vertices <math>(x_1, y_1)</math>, <math>(x_2, y_2)</math>, and <math>(x_3, y_3)</math> is given by:

$$Area\; =\; \backslash frac\{1\}\{2\}\; |x\_1(y\_2\; -\; y\_3)\; +\; x\_2(y\_3\; -\; y\_1)\; +\; x\_3(y\_1\; -\; y\_2)|$$

Area of ΔABC:

Using vertices A(-4, 5), B(0, 7), and C(5, -5):

$Area(\backslash triangle\; ABC)\; =\; \backslash frac\{1\}\{2\}\; |(-4)(7\; -\; (-5))\; +\; 0(-5\; -\; 5)\; +\; 5(5\; -\; 7)|$

$Area(\backslash triangle\; ABC)\; =\; \backslash frac\{1\}\{2\}\; |(-4)(12)\; +\; 0(-10)\; +\; 5(-2)|$

$Area(\backslash triangle\; ABC)\; =\; \backslash frac\{1\}\{2\}\; |-48\; +\; 0\; -\; 10|$

$Area(\backslash triangle\; ABC)\; =\; \backslash frac\{1\}\{2\}\; |-58|$

$Area(\backslash triangle\; ABC)\; =\; \backslash frac\{1\}\{2\}\; \backslash times\; 58\; =\; 29\; \backslash text\{\; square\; units\}$

Area of ΔACD:

Using vertices A(-4, 5), C(5, -5), and D(-4, -2):

$Area(\backslash triangle\; ACD)\; =\; \backslash frac\{1\}\{2\}\; |(-4)(-5\; -\; (-2))\; +\; 5(-2\; -\; 5)\; +\; (-4)(5\; -\; (-5))|$

$Area(\backslash triangle\; ACD)\; =\; \backslash frac\{1\}\{2\}\; |(-4)(-3)\; +\; 5(-7)\; +\; (-4)(10)|$

$Area(\backslash triangle\; ACD)\; =\; \backslash frac\{1\}\{2\}\; |12\; -\; 35\; -\; 40|$

$Area(\backslash triangle\; ACD)\; =\; \backslash frac\{1\}\{2\}\; |-63|$

$Area(\backslash triangle\; ACD)\; =\; \backslash frac\{63\}\{2\}\; \backslash text\{\; square\; units\}$

Total Area of Quadrilateral ABCD:

$Area(ABCD)\; =\; Area(\backslash triangle\; ABC)\; +\; Area(\backslash triangle\; ACD)$

$Area(ABCD)\; =\; 29\; +\; \backslash frac\{63\}\{2\}$

$Area(ABCD)\; =\; \backslash frac\{58\}\{2\}\; +\; \backslash frac\{63\}\{2\}\; =\; \backslash frac\{121\}\{2\}\; \backslash text\{\; square\; units\}$

Thus, the area of the quadrilateral is $\backslash frac\{121\}\{2\}$ square units.

Let the equilateral triangle be denoted as ΔABC, with each side having a length of 2a. So, AB = BC = CA = 2a.

We are given that the base of the triangle lies along the y-axis, and its midpoint is at the origin (0, 0). Let the base be BC.

Since the midpoint of BC is the origin O(0, 0) and BC lies on the y-axis, the coordinates of B and C must be equidistant from the origin along the y-axis. Given that the length of BC is 2a, the distance from the midpoint O to B and from O to C is 'a'.

Therefore, the coordinates of the vertices B and C are (0, a) and (0, -a) (or vice versa).

Let A be the third vertex of the triangle. Since ΔABC is equilateral, the vertex A must lie on the x-axis (as the y-axis is the perpendicular bisector of BC). Let the coordinates of A be (x, 0).

The distance from A to B (or A to C) must be equal to the side length, 2a.

Using the distance formula between A(x, 0) and B(0, a):

$AB^2\; =\; (x\; -\; 0)^2\; +\; (0\; -\; a)^2$

$(2a)^2\; =\; x^2\; +\; (-a)^2$

$4a^2\; =\; x^2\; +\; a^2$

$x^2\; =\; 4a^2\; -\; a^2$

$x^2\; =\; 3a^2$

$x\; =\; \backslash pm\; \backslash sqrt\{3\}a$

So, the possible coordinates for vertex A are ($\backslash sqrt\{3\}a$, 0) or ($-\backslash sqrt\{3\}a$, 0).

Thus, the vertices of the equilateral triangle are:

1. Base vertices: (0, a) and (0, -a)

2. Third vertex: ($\backslash sqrt\{3\}a$, 0) or ($-\backslash sqrt\{3\}a$, 0)