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). To draw the quadrilateral, we plot these points on the Cartesian plane and connect them in order: AB, BC, CD, and DA.

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

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:

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

First, let's calculate the area of ΔABC with vertices A(-4, 5), B(0, 7), and C(5, -5):

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

$=\; \backslash frac\{1\}\{2\}\; |-4(7\; +\; 5)\; +\; 0(-10)\; +\; 5(-2)|$

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

$=\; \backslash frac\{1\}\{2\}\; |-48\; -\; 10|$

$=\; \backslash frac\{1\}\{2\}\; |-58|$

$=\; \backslash frac\{1\}\{2\}\; \backslash times\; 58\; =\; 29\; \backslash text\{\; square\; units\}$

Next, let's calculate the area of ΔACD with vertices A(-4, 5), C(5, -5), and D(-4, -2):

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

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

$=\; \backslash frac\{1\}\{2\}\; |-4(-3)\; -\; 35\; -\; 4(10)|$

$=\; \backslash frac\{1\}\{2\}\; |12\; -\; 35\; -\; 40|$

$=\; \backslash frac\{1\}\{2\}\; |12\; -\; 75|$

$=\; \backslash frac\{1\}\{2\}\; |-63|$

$=\; \backslash frac\{63\}\{2\}\; \backslash text\{\; square\; units\}$

The total area of the quadrilateral ABCD is the sum of the areas of ΔABC and ΔACD:

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

$=\; 29\; +\; \backslash frac\{63\}\{2\}$

$=\; \backslash frac\{58\}\{2\}\; +\; \backslash frac\{63\}\{2\}$

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

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, <math>AB = BC = CA = 2a</math>.

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, the coordinates of B and C must be (0, a) and (0, -a) (or vice versa), because the distance from the origin to each of these points is 'a', and they lie on the y-axis.

Let B = (0, a) and C = (0, -a). The length of the base BC is the distance between these two points, which is <math>\sqrt{(0-0)^2 + (a - (-a))^2} = \sqrt{0^2 + (2a)^2} = \sqrt{4a^2} = 2a</math>, which matches the given side length.

Now, we need to find the coordinates of the third vertex, A. Since ΔABC is equilateral, the vertex A must lie on the perpendicular bisector of the base BC. The base BC lies on the y-axis, so its perpendicular bisector is the x-axis. Therefore, the y-coordinate of A must be 0. Let A = (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):

<math>AB^2 = (x - 0)^2 + (0 - a)^2 = (2a)^2</math>

<math>x^2 + (-a)^2 = 4a^2</math>

<math>x^2 + a^2 = 4a^2</math>

<math>x^2 = 4a^2 - a^2</math>

<math>x^2 = 3a^2</math>

<math>x = \pm \sqrt{3}a</math>

So, the possible coordinates for vertex A are (<math>\sqrt{3}a</math>, 0) or (<math>-\sqrt{3}a</math>, 0).

Therefore, the vertices of the equilateral triangle can be:

1. (0, a), (0, -a), and (<math>\sqrt{3}a</math>, 0)

2. (0, a), (0, -a), and (<math>-\sqrt{3}a</math>, 0)