Maths

5.

Let 0 (0, 0) be the origin and A be the mid-point of line joining P (0, –4) and B (8, 0)

Then, co-ordinate of A = $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)=\left(\frac{0+8}{2},\frac{-4+0}{2}\right)=\left(4,-2\right)$

$\therefore$ Slope of OA, m = $\frac{y_2-y_1}{x_2-x_1}=\frac{-2-0}{4-0}=-\frac{2}{4}=\frac{-1}{2}$

4.

Let A (x, 0) be the point on x-axis when is equidistant from P (7, 6) and Q (3, 4)

Then, PA = QA

Squaring both sides, we get,

$\Rightarrow x^2-14x+49+36=x^2-6x+9+16$

$\Rightarrow 49+36-9-16=14x-6x$

$\Rightarrow 60=8x$

$\Rightarrow x=\frac{60}{8}=\frac{15}{2}$

$\therefore$ The required point on x-axis is $\left(\frac{15}{2},0\right).$

3.

2. Let ABC be the equilateral triangle of side 2a and 0 be the origin. Then

AB = BC = AC = 2a

O is the mid-point of AB we have AO = a

BO = a

We know that A and B lies on y-axis so they have co-ordinate of the form (0, y).

Hence, co-ordinate of A is (0, a) and that of B is (0, –a)

Since OC, bisects AB at right angle, by Pythagoras theorem,

AC²= OA² + OC²

$\Rightarrow {\left(2a\right)}^2={\left(a\right)}^2+OC{}^2$

$\Rightarrow OC{}^2=4a^2-a^2$

And as C we on x-axis it has co-ordinate of the form (x, 0)

Exercise 9.1

1. Let the given points be A(–4, 5), B(0, 7), C(5, –5) and D(–4, –2).

Then quadrilated ABCD can be plotted on the graph by joining the points A, B, C and D.

We connect diagonal AC such that

area (ABCD) = (ΔABC) + (ΔADC)

Now,

$\left(\Delta ABC\right)=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$

$=\frac{1}{2}\left|-4\left[7-\left(-5\right)\right]+0\left[-5-5\right]+5\left[5-7\right]\right|\text{\hspace{0.17em}unit}{}^2$

$=\frac{1}{2}\left|-4\left(7+5\right)+0+5\left(5-7\right)\right|\text{\hspace{0.17em}unit}{}^2$

$=\frac{1}{2}|\left(-4\right)*12+5*\left(-2\right)|unit{}^2$

$=\frac{1}{2}|-48-10|unit{}^2$

$=\frac{1}{2}\left|-58\right|\text{\hspace{0.17em}unit}{}^2$

$=\frac{58}{2}\text{unit\hspace{0.17em}}{}^2=29\text{\hspace{0.17em}unit}{}^2$

Similarly, $\left(\Delta ACD\right)=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$

$=\frac{1}{2}\left|-4\left[-5-\left(2\right)\right]+5\left[-2-5\right]+\left(-4\right)\left[5-\left(-5\right)\right]\right|\text{\hspace{0.17em}unit}{}^2$

$=\frac{1}{2}|-4\left(-5+2\right)+5\left(-2-5\right)-4\left(5+5\right)|unit{}^2$

$=\frac{1}{2}|\left(-4\right)*\left(-3\right)+5*\left(-7\right)-4*\left(10\right)|unit{}^2$

$=\frac{1}{2}|12-35-40|unit{}^2$

$=\frac{1}{2}\left|-63\right|\text{\hspace{0.17em}unit}{}^2$

$=\frac{63}{2}\text{\hspace{0.17em}unit}{}^2$

Hence, area (ABCD) = $29+\frac{63}{2}\text{\hspace{0.17em}unit}{}^2$

$=\frac{58+63}{2}\text{\hspace{0.17em}unit}{}^2=\frac{121}{2}\text{\hspace{0.17em}unit}{}^2$

students can check the table for the principal values for all ITFs below;

To understand this, Assume you have a bucket that has infinite number of apples and if your mother asks "give me the apple". How will you figure out which one is "The Apple", she is asking for.

Similarly any inversre trigonometric functions behaves like a Many-one Function; which means,

For Example $\sin^ {-1}\left (\frac {2} {3}\right)$ can have many solutions, we need to fix one solutions which can be used as standerd value for the function.

• A standerd value (Angles) of any inverse trigonometric value lies between a fiexed range is known as principal value. 

For Ex; The value of $\sin^ {-1}\left (\frac {2} {3}\right)$ will always lie between –? /2 to? /2.

$y={\sin }^{?1}\left(\frac{2}{3}\right)?\sin \left(y\right)=\frac{2}{3}?y?[?\frac{?}{2},\frac{?}{2}]$  

The Class 12 Relations and Functions explores various types of Functions, Students can check main types dicussed in this chapter below;

• One-One Function (Injective)

• Onto Function (Surjective)

• One-One and Onto Function (Bijective)

• Identity Function

• Constant Function

• Inverse of a Function

• Composite Functions

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