Question 14 Marks
Show that the quadrilateral with vertices $(3, 2), (0, 5), (- 3, 2)$ and $(0, -1)$ is a square.
AnswerLet $A(3, 2), B(0, 5), C(-3, 2)$ and $D(0, -1)$ are the verticles of quadrilateral.
$\text { Now } A B=\sqrt{(3-0)^2+(2-5)^2}$
$=\sqrt{18} \text { units. }$
$\Rightarrow A B^2=18$
$B C=\sqrt{(0+3)^2+(5-2)^2}$
$=\sqrt{18} \text { units. }$
$\Rightarrow B C^2=18$ $C D=\sqrt{(-3-0)^2+(2+1)^2}$
$=\sqrt{18} \text { units. }$
$\Rightarrow C D^2=18$
$\text { Also } A D=\sqrt{(3-0)^2+(2+1)^2}$
$=\sqrt{18} \text { units. }$
$\Rightarrow A D^2=18 .$
$\text { Here } A B=B C=C D=D A$
$=\sqrt{18} \text { units. }$
Also
$A C^2=(3+3)^2+(2-2)^2$
$A C^2=36$
or
$BD ^2=(0-0)^2+(5+1)^2$
$=36 $
$\Rightarrow \text { Diagonal } AC = BD$
Hence, $A B C D$ is a square.
Hence proved.
View full question & answer→Question 24 Marks
Show that each of the triangles whose vertices are given below are isosceles :
$(i)\ (8, 2), (5,-3)$ and $(0,0)$
$(ii)\ (0,6), (-5, 3)$ and $(3,1).$
Answer$(i)$ Let $A (8, 2) B (5, -3)$ and $C (0, 0)$ be the given point
Then $AB = \sqrt{(8-5)^2+(2-3)^2}$
$=\sqrt{9+25}=\sqrt{34} \text { units. } $
$ B C=\sqrt{(5-10)^2+(-3-0)^2} $
$=\sqrt{25+9}=\sqrt{34} \text { units. } $
$ A C=\sqrt{(8-0)^2+(2-0)^2}$
$ =\sqrt{64+4}=\sqrt{68} . \text { units. } $
$ \text { Here } A B=B C=\sqrt{34} .$
Hence, the triangle is isosceles.
$(ii)$ The given points are $A(0, 6), B(-5, 3)$ and $C(3, 1)$
$\text { Then } A B=\sqrt{(0+5)^2+\left(6-3^2\right)} $
$=\sqrt{25+9}=\sqrt{34} \text { units. } $
$B C=\sqrt{(-5-3)^2+(3-1)^2} $
$=\sqrt{64+4}=\sqrt{68} \text { units. }$
Also $AC =\sqrt{(0-3)^2+(6-1)^2}$
$=\sqrt{9+25}=\sqrt{34}$ units.
Since $A B=A C=\sqrt{34}$
Hence, the triangle is an isosceles.
Hence proved.
View full question & answer→Question 34 Marks
By using the distance formula prove that each of the following sets of points are the vertices of a right angled triangle.
(i) $(6, 2), (3, -1)$ and $(- 2, 4)$
(ii) $(-2, 2), (8, -2)$ and $(-4, -3).$
AnswerLet $A(6, 2), B(3, -1)$ and $C(-2, 4)$ be the given points
$A B=\sqrt{(6-3)^2+(2+1)^2}$
$=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2} \text { units. }$
$B C=\sqrt{(3+2)^2+(-1-4)^2}$
$=\sqrt{25+25}=5 \sqrt{2} \text { units. }$
$A C=\sqrt{(6+2)^2+(2-4)^2}$
$=\sqrt{64+4}$
$=\sqrt{68} \text { units. }$
$\text { Now } A B^2+B C^2=(3 \sqrt{2})^2+(5 \sqrt{2})^2$
$=18+50=68=A C^2$
$A B^2+B C^2=A C^2$
Hence, the triangle is right angled at A.
Hence proved.
(ii) Let $A (-2, 2), B(8, -2)$ and $C(-4, -3)$ be the given points
$A B=\sqrt{(8+2)^2+(-2-2)^2}$
$=\sqrt{116} \text { units. }$
$B C=\sqrt{(8+4)^2+(-2+3)^2}$
$=\sqrt{145} \text { units. }$
$A C=\sqrt{(-2+4)^2+(2+3)^2}$
$=\sqrt{29} \text { units. }$
$\text { Now } A B^2+A C^2$
$=116+29$
$=145$
$=B C^2$
Since $AB + AC^2 = BC^2$
Hence, the triangle is right angled at $A.$
View full question & answer→Question 44 Marks
If the image of the point $(2,1)$ with respect to the line mirror be $(5, 2).$ Find the equation of the mirror.
AnswerLet $CD$ be the line mirror with slope $m_1.$
Now the slope of the line joining $A(2, 1)$ and $B(5, 2).$
$m_2 = \frac{2-1}{5-2}=\frac{1}{3}$
Since $CD \perp AB$
So, $m_1m_2 = -1$
$\Rightarrow m_1 \times \frac{1}{3}=-1$
$\Rightarrow m_1=-3 .$
Now mid point of $A B=\left(\frac{2+5}{2}, \frac{1+2}{2}\right)=\left(\frac{7}{2}, \frac{3}{2}\right)$
Equation of the mirror $CD,$
$y = y_1 = m(x - x_1)$
$\Rightarrow y-\frac{3}{2}=-3\left(x-\frac{7}{2}\right)$
$\Rightarrow y-\frac{3}{2}=-3 x+\frac{21}{2}$
$\Rightarrow 2 y-3=-6 x+21$
$\Rightarrow 6 x+2 y-3-21=0$
$\Rightarrow 6 x+2 y-24=0$
$\text { or } 3 x+y-12=0 .$ View full question & answer→Question 54 Marks
Find a general equation of a line which passes through:
$(i) (0, -5)$ and $(3, 0) (ii) (2, 3)$ and $(-1, 2).$
AnswerWe have the equation of a line which passes through $(x_1,y_1)$ and $(x_2,y_2)$ is
$y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$
(i) Putting $x_1 = 0, y_1 = -5$ and $x_2 = 3, y_2 = 0$
$y-(-5)=\frac{0-(-5)}{3-0}(x-0)$
$\Rightarrow y+5=\frac{5}{3}(x-0)$
$\Rightarrow 3 y+15=5 x$
$\Rightarrow 5 x-3 y-15=0$
Which is the required equation.
(ii) Putting$ x_1 = 2, y_1 = 3$ and $x = -1, y_2 = 2$
$y-3=\frac{2-3}{-1-2}(x-2)$
$\Rightarrow y-3=\frac{-1}{-3}(x-2)$
$\Rightarrow 3 y-9=(x-2)$
$\Rightarrow x-2-3 y+9=0$
$\Rightarrow x-3 y+7=0$
which is the equation of the required line.
$\Rightarrow 3y - 9 = (x - 2)$
$\Rightarrow x - 2 - 3y + 9 = 0$
$\Rightarrow x - 3y + 7 = 0$
which is the equation of the required line.
View full question & answer→Question 64 Marks
The figure alongside $($not drawn to scale$)$ represents the lines $y = x + 1$ and $y = \sqrt{3} x-1$.
$(i)$ Find the angle which the line $y = x + 1$ makes with $X-$ axis.
$(ii)$ Find the angle which the line $y = \sqrt{3} x-1$ makes with $X-$ axis.
$(iii)$ Determine angle $\theta $.
$(iv)$ Find the point where the line $y = x + 1$ meets $X-$ axis.
$(v)$ Find the point where the line $y = \sqrt{3} x-1$ meets $Y-$ axis. Answer$\text { (i) } y=x+1$
$\Rightarrow m_1=\tan \theta_1=1=\tan 45^{\circ}$
$\Rightarrow \theta_1=45^{\circ} .$
$\text { (ii) } y=\sqrt{3} x-1$
$\Rightarrow m_2=\tan \theta_2=\sqrt{3}=\tan 60^{\circ}$
$\Rightarrow \theta_2=60^{\circ}$
$(iii) \therefore \theta=\theta_2-\theta_1$.
$( \because$ Exterior $\angle=$ Sum of interior opposite $\angle s )$
$=60^{\circ}-45^{\circ}=15^{\circ} \text {. }$
$(iv) $ Put $y=0$ in $y=x+1$, we get
$0=x+1$
$\Rightarrow x=-1$
$\therefore$ The required points is $(-1,-0)$.
$(v)$ Put $x =0$ in $y =\sqrt{3} x-1$, we get $y =-1$
$\therefore$ The required point is $(0,-1)$.
View full question & answer→Question 74 Marks
The vertices of a triangle are $A(10, 4), B(- 4, 9)$ and $C(- 2, -1)$. Find the
AnswerLet $\text{AD, BE}$ and $CF$ be the three altitudes of $\triangle ABC$ then
$AD \perp BC$
$BE \perp CA$
and $CF \perp AB.$
Slope of $B C=\frac{-1-9}{-2+4}=-5$
Since $A D \perp B C$
Slope of $B C \times$ slope of $A D=-1$
Slope of $A D=\frac{-1}{-5}=\frac{1}{5}$
Therefore $AD \perp BC$
Since$, AD$ passes through $A(10, 4)$
So$,$ equation of $AD$ is
$y - y_1= m (x - x_1)$
$y-4=\frac{1}{5}(x-10)$
$\Rightarrow 5y - 20 = x - 10$
$\Rightarrow x - 5y + 10 = 0 ...(i)$
Now $,$ Slope of $AC = \frac{4+1}{10+2}=\frac{-5}{12}$
since$ BE \perp AC$
Slope of $BE\ x$ Slope of $AC = -1$
So$,$ Slope of $BE = \frac{-1 \times 12}{5}=-\frac{12}{5}$.
Equation of $BE$ which passes through $B(-4, 9)$ is
$y - y_1 = m(x - x_1)$
$y-9=\frac{-12}{5}(x+4)$
or $12x + 5y + 3 = 0 ...(ii)$
Slope of $AB\ x$ Slope of $CF = -1$
$\Rightarrow-\frac{5}{14} \times$ Slope of $CF =-1$
$\Rightarrow$ Slope of $CF =\frac{14}{5}$
Equation of $CF$ which passes through $C(-2, -1)$ is
$y - y_1 = m (x - x_1)$
$y+1=\frac{14}{5}(x+2)$
$\Rightarrow 14x - 5y + 23 = 0 ...(iii)$
Thus, the equation of altitudes of $\triangle \text{ABC}$ are
$x - 5y + 10 = 0$
$12x + 5y + 3 = 0$
and $14x - 5y + 23 = 0.$ View full question & answer→Question 84 Marks
Find the value of $'a\ ’$ for which the following points $A (a, 3), B (2, 1)$ and $C (5, a)$ are collinear. Hence find the equation of the line.
AnswerEquation of the line passing through $AC$ is
$(y-3)=\left(\frac{a-3}{5-a}\right)(x-a)$
As if $A, B$ and $C$ are callinear than $B$ will satisfy it, $i.e.,$
$(1-3)=\left(\frac{a-3}{5-a}\right)(2-a)$
$-2(5-a)=(a-3)(2-a)$
$-10+2 a=2 a-6-a^2+3 a$
$a^2-3 a-4=0$
$a^2-4 a+a-4=0$
$a(a-4)+1(a-4)=0$
$(a-4)(a+1)=0$
$\Rightarrow a=4 \text { or }-1.$
Thus, required equation of straight line is
$(y-3)=\left(\frac{4-3}{5-4}\right)(x-4)$
$y-3=\left(\frac{1}{1}\right)(x-4)$
$x-y-1=0$
or
$(y-3)=\left(\frac{-1-3}{5+1}\right)(x+1)$
$(y-3)=\left(-\frac{4}{6}\right)(x+1)$
$y-3=\frac{-2}{3}(x+1)$
$3y - 9 - 2x - 2$
$2x + 3y - 7 = 0.$ View full question & answer→Question 94 Marks
Find the image of a point $(-1, 2)$ in the line joining $(2, 1)$ and $(- 3, 2)$.
AnswerLet $D(\alpha , \beta)$ be the image of point $C(-1, 2)$ in the line joining the points $A(2, 1)$ and $B(-3, 2)$.
Since $AB$ is the perpendicular bisector of $CD$.
So, Slope of $AB\ x$ Slope of $CD = -1$
$\Rightarrow \frac{2-1}{-3-2} \times \frac{\beta-2}{\alpha+1}=-1$
$\Rightarrow \frac{1}{-5} \times \frac{\beta-2}{\alpha+1}=-1$
$\Rightarrow \beta-2=5 \alpha+5$
$\Rightarrow 5\alpha - \beta + 7 = 0 ...(i)$
Equation of line $A B,$
$y-1=\frac{2-1}{-3-2}(x-2)$
$\Rightarrow y-1=\frac{1}{-5}(x-2)$
$\Rightarrow-5(y-1)=x-2$
$\Rightarrow x-2+5 y-5=0$
$\Rightarrow x + 5y - 7 = 0 ...(ii)$
Since, midpoint of $C D\ (\frac{\alpha-1}{2},\frac{\beta+2}{2})$ lies on $A B$.
$\frac{\alpha-1}{2}+5\left(\frac{\beta+2}{2}\right)-7=0$
$\Rightarrow \alpha-1+5 \beta+10-14=0$
$\Rightarrow \alpha+5 \beta-5=0 \ldots \text { (iii) }$
Solving $(i)$ and $(iii),$ we get
$\alpha=\frac{-15}{13} $ and $ \beta=\frac{16}{13}$
Hence, coordination of $D$ are $\left(-\frac{15}{13}, \frac{16}{13}\right)$. View full question & answer→Question 104 Marks
Determine the centre of the circle on which the points $(1, 7), (7 – 1),$ and $(8, 6)$ lie. What is the radius of the circle?
AnswerLet $P(x, y)$ be the centre of the circle and $A(1, 7), B(7, -1)$ and $C(8, 6)$ be the given points.
Then $PA = PB = PC =$ radius
$\Rightarrow PA^2 = PB = PC = r^2$
$\Rightarrow (x - 1)^2 + y( - 7)^2 = r^2 ...(i)$
$(x - 7)^2 + (y + 1)^2= r^2 ...(ii)$
Also $(x - 8)^2 + (y - 6)^2 = r^2 ...(iii)$
Subtracting $(ii)$ from $(i)$
$(x^2 + 1 - 2x + y^2 + 49 - 14y)$
$-(x^2 + 49 - 14x + y^2 + 1 + 2y) = 0$
$\Rightarrow 12x - 16y = 0$
$y=\frac{3}{4} x ...(iv)$
Subtracting $(iii)$ from $(ii)$
$\Rightarrow\left(x^2+49-14 x+y^2+1+2 y\right)$
$-\left(x^2+64-16 x+y^2+36-12 y\right)=0$
$\Rightarrow 2 x+14 y-50=0$
$\Rightarrow x+7\left(\frac{3}{4} x\right)-25=0$
$\Rightarrow \frac{25 x}{4}=25$
$\Rightarrow x =4$
From $(iv) y=\frac{3}{4} x=\frac{3}{4} \times 4$
$y = 3.$
The centre is $P(4, 3).$
Also radius $r=P A=\sqrt{(4-1)^2+(3-7)^2}$
$=\sqrt{9+16}$
$=5$ units View full question & answer→