Maths

8. Let P (x, –1), Q (2, 1) and R (4, 5) be the collinear points. Then,

Slope of PQ = Slope of QR

$\Rightarrow \frac{1-\left(-1\right)}{2-x}=\frac{5-1}{4-2}$

$\Rightarrow \frac{1+1}{2-x}=\frac{4}{2}$

$\Rightarrow 2*2=4\left(2-x\right)$

$\Rightarrow 4=8-4x$

$\Rightarrow x=\frac{4}{4}$

$\Rightarrow x=1$

7. 

Let l be the line making 30° with y-axis as shown in figure. Then,

Angle a = $\angle b$ + 90° (Sum of exterior angle of a triangle)

$\Rightarrow \angle a=30°+90°$  ( $30°=\angle b$ vertically opposite angle)

$\Rightarrow \angle a=120°$

So, slope of line l = tan a

= m = tan 120°

= tan (180° – 60°)

= –tan 60°

$=-\surd 3$

6.

. Let the given point be A (4, 4), B (3, 5) and C (–1, –1)

Then, slope of AB, m₁ = $\frac{5-4}{3-4}=\frac{1}{-1}=-1$

Slope of AC, m₂ = $\frac{-1-4}{-1-4}=\frac{-5}{-5}=1$

And slope of BC, m₃ = $=\frac{-1-5}{-1-3}=\frac{-6}{-4}=\frac{3}{2}$

As m₁ – m₂ = –1 * 1 = –1

$\Rightarrow m_1=\frac{-1}{m_2}$

We conclude that AB and AC are perpendicular to each other.

Hence, ABC is a right-angle triangle right-angled at A

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;