Class 11 Maths NCERT Solutions Chapter 9 Ex 9.1

Let the vertices be A(-4,5), B(0,7), C(5,-5) and D(-4,-2), plotted and joined in order as shown below.

To find the area, draw diagonal AC, which splits quadrilateral ABCD into \triangle ABC and \triangle ACD.

Area of △ABC

Using the area formula with A(-4,5), B(0,7), C(5,-5):

\text{Area}=\dfrac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|

=\dfrac{1}{2}\left|(-4)(7-(-5))+0((-5)-5)+5(5-7)\right|

=\dfrac{1}{2}\left|(-4)(12)+0(-10)+5(-2)\right|=\dfrac{1}{2}\left|-48+0-10\right|=\dfrac{1}{2}(58)=29

Area of △ACD

Using the area formula with A(-4,5), C(5,-5), D(-4,-2):

=\dfrac{1}{2}\left|(-4)((-5)-(-2))+5((-2)-5)+(-4)(5-(-5))\right|

=\dfrac{1}{2}\left|(-4)(-3)+5(-7)+(-4)(10)\right|=\dfrac{1}{2}\left|12-35-40\right|=\dfrac{1}{2}(63)=\dfrac{63}{2}

Area of quadrilateral ABCD

=\text{Area}(\triangle ABC)+\text{Area}(\triangle ACD)=29+\dfrac{63}{2}=\dfrac{58+63}{2}=\dfrac{121}{2}

Since the base lies along the y-axis with length 2a and its mid-point is at the origin, the two base vertices are equidistant from the origin along the y-axis. Let them be:

B(0,a) and C(0,-a)

Let the third vertex (apex) be A(h,0), which by symmetry must lie on the x-axis. Since \triangle ABC is equilateral, AB=2a. Using the distance formula:

AB=\sqrt{(h-0)^2+(0-a)^2}=2a

Squaring both sides:

h^2+a^2=4a^2\;\Rightarrow\;h^2=3a^2\;\Rightarrow\;h=\pm\sqrt{3}\,a

The distance between P and Q is always given by the distance formula:

PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

(i) When PQ is parallel to the y-axis

Every point on a line parallel to the y-axis has the same x-coordinate, so x_1=x_2. Substituting in the distance formula:

PQ=\sqrt{0+(y_2-y_1)^2}=|y_2-y_1|

(ii) When PQ is parallel to the x-axis

Every point on a line parallel to the x-axis has the same y-coordinate, so y_1=y_2. Substituting in the distance formula:

PQ=\sqrt{(x_2-x_1)^2+0}=|x_2-x_1|

Let the required point on the x-axis be P(x,0). Since P is equidistant from A(7,6) and B(3,4):

PA=PB\;\Rightarrow\;PA^2=PB^2

(x-7)^2+(0-6)^2=(x-3)^2+(0-4)^2

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

-14x+85=-6x+25

-14x+6x=25-85\;\Rightarrow\;-8x=-60\;\Rightarrow\;x=\dfrac{15}{2}

Mid-point of PB, using P(0,-4) and B(8,0):

M=\left(\dfrac{0+8}{2},\dfrac{-4+0}{2}\right)=(4,-2)

The required line passes through the origin O(0,0) and M(4,-2). Its slope is:

m=\dfrac{-2-0}{4-0}=-\dfrac{2}{4}=-\dfrac{1}{2}

Let A(4,4), B(3,5) and C(-1,-1). Instead of using the Pythagoras theorem, we show that two of the sides are perpendicular by checking that the product of their slopes is -1.

Slope of AB:

m_{AB}=\dfrac{5-4}{3-4}=\dfrac{1}{-1}=-1

Slope of AC:

m_{AC}=\dfrac{-1-4}{-1-4}=\dfrac{-5}{-5}=1

Product of slopes:

m_{AB}\times m_{AC}=(-1)(1)=-1

Since m_{AB}\times m_{AC}=-1, the lines AB and AC are perpendicular, so the angle at vertex A is 90^\circ.

The slope of a line is m=\tan\theta, where \theta is the inclination — the angle the line makes with the positive direction of the x-axis, measured anticlockwise. Here, the angle is given with respect to the positive y-axis, so it must first be converted to the inclination.

Measuring 30^\circ anticlockwise from the positive y-axis (which itself is at 90^\circ from the positive x-axis) gives an inclination of:

\theta=90^\circ+30^\circ=120^\circ

Therefore, the slope of the line is:

m=\tan120^\circ=\tan(180^\circ-60^\circ)=-\tan60^\circ=-\sqrt{3}

Let A(-2,-1), B(4,0), C(3,3) and D(-3,2). Instead of using the distance formula, we show that both pairs of opposite sides are parallel, using slopes.

Slope of AB:

m_{AB}=\dfrac{0-(-1)}{4-(-2)}=\dfrac{1}{6}

Slope of DC:

m_{DC}=\dfrac{3-2}{3-(-3)}=\dfrac{1}{6}

Since m_{AB}=m_{DC}, AB\parallel DC.

Slope of BC:

m_{BC}=\dfrac{3-0}{3-4}=\dfrac{3}{-1}=-3

Slope of AD:

m_{AD}=\dfrac{2-(-1)}{-3-(-2)}=\dfrac{3}{-1}=-3

Since m_{BC}=m_{AD}, BC\parallel AD.

Slope of the x-axis: m_1=0.

Slope of the line joining (3,-1) and (4,-2):

m_2=\dfrac{-2-(-1)}{4-3}=\dfrac{-1}{1}=-1

Using the formula for the acute angle between two lines:

\tan\theta=\left|\dfrac{m_2-m_1}{1+m_1m_2}\right|=\left|\dfrac{-1-0}{1+0}\right|=|-1|=1

\theta=\tan^{-1}(1)=45^\circ

Let the slope of one line be m, so the slope of the other line is 2m. Using the formula for the angle between two lines:

\tan\theta=\left|\dfrac{m_2-m_1}{1+m_1m_2}\right|

With m_1=m, m_2=2m and \tan\theta=\dfrac{1}{3}:

\dfrac{1}{3}=\left|\dfrac{2m-m}{1+2m^2}\right|=\left|\dfrac{m}{1+2m^2}\right|

Since 1+2m^2 is always positive, this gives two cases.

Case 1: m is positive

\dfrac{m}{1+2m^2}=\dfrac{1}{3}\;\Rightarrow\;3m=1+2m^2\;\Rightarrow\;2m^2-3m+1=0

(2m-1)(m-1)=0\;\Rightarrow\;m=\dfrac{1}{2}\text{ or }m=1

Case 2: m is negative

-\dfrac{m}{1+2m^2}=\dfrac{1}{3}\;\Rightarrow\;-3m=1+2m^2\;\Rightarrow\;2m^2+3m+1=0

(2m+1)(m+1)=0\;\Rightarrow\;m=-\dfrac{1}{2}\text{ or }m=-1

Each value of m gives a pair of slopes (m,2m):

The line passes through the two points (x_1,y_1) and (h,k). By definition, the slope of a line through two points is the ratio of the difference in y-coordinates to the difference in x-coordinates:

m=\dfrac{k-y_1}{h-x_1}

Multiplying both sides by (h-x_1):

m(h-x_1)=k-y_1

That is:

k-y_1=m(h-x_1)