Question 1015 Marks
Show that the area of the triangle formed by the lines $y=m_1 x, y=m_2 x$ and $y=c$ is equal to $\frac{c^2}{4}(\sqrt{33}+\sqrt{11})$, where $m_1$, $m _2$ are the roots of the equation $x ^2+(\sqrt{3}+2) x +\sqrt{3}-1=0$.
Answer$y=m_1 x, y=m_2 x$ and $y=c$
Vertices of triangle formed by above lines are
$\text{A}(0,0) ; \ \text{B}(\frac{\text{c}}{\text{m}_1},\text{c}); \ \text{C}(\frac{\text{c}}{\text{m}_2},\text{c})$
So Area of triangle when three vertices are given is
$\frac{1}{2}(\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2))$
$=\frac{1}{2}\bigg[\bigg|\frac{\text{c}^2}{\text{m}_1}-\frac{\text{c}^2}{\text{m}_2}\bigg|\bigg]=\frac{\text{c}^2}{2}\bigg[\bigg|\frac{\text{m}_2-\text{m}_1}{\text{m}_1\text{m}_2}\bigg|\bigg]$
Given $m_1$and $m_2$are roots of $\text{x}^2+(\sqrt3+2)\text{x}+\sqrt3-1=0$
Product of roots $=\text{m}_1\text{m}_2=\sqrt3-1$
$|\text{m}_2-\text{m}_1|=\sqrt{(\text{m}_2+\text{m}_1)^2-4\text{m}_1\text{m}_2}=\sqrt{(\sqrt3+2)^2-4\sqrt3+4}$
$|\text{m}_2-\text{m}_1|=\sqrt{3+4+4\sqrt3-4\sqrt3+4}=\sqrt{11}$
Area $=\frac{\text{c}^2}{2}\bigg[\frac{\sqrt11}{\sqrt3-1}\bigg]$
Rationalising denominator gives $\frac{\text{c}^2}{4}\big[\sqrt33+\sqrt11\big]$
Hence proved
View full question & answer→Question 1025 Marks
Find the equation of the straight line which passes through the point of intersection of the lines 3x - y = 5 and x + 3y = 1 and makes equal and positive intercepts on the axes.
AnswerThe required line is
$(3\text{x}-\text{y}-5)+\lambda(\text{x}+3\text{y}-1)=0$
or, $(3+\lambda)\text{x}+(-1+3\lambda)\text{y}-5-\lambda=0$
or, $\frac{\text{x}}{\Big(\frac{5+\lambda}{3+\lambda}\Big)}+\frac{\text{y}}{\frac{5+\lambda}{3\lambda-1}}=1$
And the line makes equal and positive intercepts with the line (given)
$\therefore \ \frac{5+\lambda}{3+\lambda}=\frac{5+\lambda}{3\lambda-1}$
$3\lambda-1=3+\lambda$
$2\lambda=4$
$\lambda=2$
$\therefore$ The required line is
3x - y - 5 + 2x + 6y - 2 = 0
or, 5x + 5y = 7
View full question & answer→Question 1035 Marks
Prove that the following sets of three lines are concurrent:
15x - 18y + 1 = 0, 12x + 10y - 3 = 0 and 6x + 66y - 11 = 0
AnswerIf the lines are concurrent then point of intersection of any two lines satisfies the third line
15x - 18y + 1 = 0 ...(1)
12x + 10y - 3 = 0 ...(2)
6x + 66y - 11 = 0 ...(3)
Solving (1) and (2)
$\text{x}=\frac{18\text{y}-1}{5}$
$12\Big(\frac{18\text{y}-1}{15}\Big)+10\text{y}-3=0$
$216\text{y}-12+150\text{y}-45=0$
$366\text{y}=57$
$\text{y}=\frac{57}{366}=\frac{19}{122}$
$\Rightarrow\text{x}=\frac{18\text{y}-1}{15}$
$=\frac{18\times\frac{19}{122}-1}{15}$
$=\frac{18\times19-122}{122\times15}$
$=\frac{342-122}{1730}$
$=\frac{220}{1730}$
$=\frac{22}{173}$
Putting x and y in (3)
$6\Big(\frac{22}{173}\Big)+66\Big(\frac{19}{122}\Big)-11=0$
$6\times22\times122+66\times19\times173-11\times173\times122=0$
$0=0$
View full question & answer→Question 1045 Marks
Find the equations of the lines through the point of intersection of the lines x - 3y + 1 = 0 and 2x + 5y - 9 = 0 and whose distance from the origin is $\sqrt{5}.$
AnswerThe required line is
$\text{x}-3\text{y}+1+\lambda(2\text{x}+5\text{y}-9)=0$
or, $(1+2\lambda)\text{x}+(-3+5\lambda)\text{y}+1-9\lambda=0$
Distance from origin of this line is
$\Bigg|\frac{(1+2\lambda)0+(-3+5\lambda)0+1-9\lambda}{\sqrt{(1+2\lambda)^2+(5\lambda-3)^2}}\Bigg| \ \Big[\text{using} \ \frac{\text{ax}_1+\text{by}_1+\text{c}}{\sqrt{\text{a}^2+\text{b}^2}}\Big]$
$\Rightarrow\sqrt{5}=\Big|\frac{1-9\lambda}{\sqrt{1+4\lambda^2+4\lambda+25\lambda^2+9-30\lambda}}\Big|$
$\Rightarrow\sqrt{5}=\Big|\frac{1-9\lambda}{\sqrt{10+29\lambda^2-26\lambda}}\Big|$
$\Rightarrow5(10+29\lambda^2-26\lambda)=(1-9\lambda)^2$
$\Rightarrow50+145\lambda^2-130\lambda=1+81\lambda^2-18\lambda^2$
$\Rightarrow64\lambda^2-112\lambda+49=0$
$\Rightarrow(8\lambda-7)^2=0$ or, $\lambda=\frac{7}{8}$
$\therefore$ Required line is
$\text{x}-3\text{y}+1+\frac{7}{8}(2\text{x}+5\text{y}-9)=0$
$\Rightarrow8\text{x}-24\text{y}+8+14\text{x}+35\text{y}-63=0$
$\Rightarrow22\text{x}+11\text{y}-55=0$
$\Rightarrow2\text{x}+\text{y}-5=0$
View full question & answer→Question 1055 Marks
Find the equation of straight line passing through (-2, -7) and having an intercept of length 3 between the straight lines 4x + 3y = 12 and 4x + 3y = 3.
AnswerThe equation of any line passing through (-2, -7) is$\frac{\text{x}+2}{\cos\theta}_+\frac{\text{y}+7}{\sin\theta}=\text{r}$
B and C are at distance r and (r + 3)
Thus, coordinates of B and C are $(-2+\text{r}\cos\theta,\ -7+\text{r}\sin\theta)$ and $(-2 + (\text{r} + 3)\cos\theta,\ -7+(\text{r}+3)\sin\theta)$
B lies on 4x + 3y = 12
$\Rightarrow4(-2+\text{r}\cos\theta)+3(-7+\text{r}\sin\theta)=12\dots(1)$
C lies on 4x + 3y = 3
$\Rightarrow4(-2+(\text{r}+3)\cos\theta)+3(-7+(\text{r}+3)\sin\theta)=3\dots(2)$
Subtarcting (1) from (2)
$12\cos\theta +9\sin\theta=-9$
$\Rightarrow4\cos\theta=-3(1-\sin\theta)$
$\Rightarrow16\cos^2\theta=9(1+\sin^2\theta-2\sin\theta)$
$\Rightarrow16(1-\sin^2\theta)=9(1+\sin^2\theta-2\sin\theta)$
$\Rightarrow16-16\sin^2\theta =9+9\sin^2\theta-18\sin\theta$
$\Rightarrow25\sin^2\theta-18\sin\theta-7=0$
$\Rightarrow25\sin^2\theta-25\sin\theta+7\sin\theta-7=0$
$\Rightarrow25\sin\theta(\sin\theta-1)-7(\sin\theta-1)=0$
$\sin\theta=1,\ \sin\theta=\frac{7}{25}$
Now, $\sin\theta=1\Rightarrow\cos\theta=0$
$\therefore\text{x}+2=0\dots(1)$
and if, $\sin\theta=\frac{7}{25}$ then $\cos\theta=\frac{24}{5}$
$\therefore\frac{\text{x}+2}{\frac{24}{25}}=\frac{\text{y}+7}{\frac{7}{25}}$
$\Rightarrow7\text{x}+\text{24y}+182=9\dots(2)$
View full question & answer→Question 1065 Marks
The equations of perpendicular bisectors of the sides $A B$ and $A C$ of a triangle $A B C$ are $x-y+5=0$ and $x+2 y=0$ respectively. If the point $A$ is $(1,-2)$, find the equation of the line $B C$.
AnswerLet $\left( x _1, y _1\right)$ and $\left( x _2, y _2\right)$ be the coordinates of B and C .
Perpendicular bisector of $A B$ is $x-y+5=0$
Its slope = 1
Coordinates of F $\Big(\frac{\text{x}_1+1}{2},\frac{\text{y}_1-2}{2}\Big)$
F lies on the x - y + 5 = 0
$\Rightarrow\frac{\text{x}_1+1}{2}-\frac{\text{y}_1-2}{2}+5=0$
$\Rightarrow\text{x}_1+1-\text{y}_1+2+10=0$
$\text{x}_1-\text{y}_1+13=0 \ ...(1)$
AB is perpendicular to HF
(Slope of AB)(Slope of HF) = -1
$\Big(\frac{\text{y}_1+2}{\text{X}_1-1}\Big)(1)=-1$
$\text{x}_1+\text{y}_1+1=0 \ ...(2)$
Solving equation (1) and (2),
$x_1 = -7, y_1 = 6$
Thus, B is (-7, 6)
Now, perpendicular bisector of AC is
x + 2y = 0
Slope of this is $=-\frac{1}{2}$
Mid-point of $\text{ACE}=\Big(\frac{\text{x}_2+1}{2},\frac{\text{y}_2-2}{2}\Big)$
E lies on perpendicular bisector of AC
$\Rightarrow\Big(\frac{\text{x}_2+1}{2}\Big)+2\Big(\frac{\text{y}_2-2}{2}\Big)=0$
$\text{x}_2+1+2\text{y}_2-4=0$
$\text{x}_2+2\text{y}_2-3=0 \ ...(3)$
AC is perpendicular to HE
(Slope of AC)(Slope of HE) = -1
$\Big(\frac{\text{y}_2+2}{\text{x}_2-1}\Big)\Big(-\frac{1}{2}\Big)=-1$
$\text{y}_2+2=2\text{x}_2-2$
$2\text{x}_2-\text{y}_2=4 \ ...(4)$
Solving equation (3) and (4), we get
$\text{x}_2=\frac{11}{5},\text{y}_2=\frac{2}{5}$
Thus, point C is $\Big(\frac{11}{5},\frac{2}{5}\Big)$
Equation of BC is
$\text{y}-\text{y}_1=\frac{\text{y}_2-\text{y}_1}{\text{x}_2-\text{x}_1}(\text{x}-\text{x}_1)$
$\text{y}-6=\frac{\frac{2}{5}-6}{\frac{11}{5}+7}(\text{x}+7)$
$\text{y}-6=\frac{-\frac{28}{5}}{\frac{46}{5}}(\text{x}+7)$
$\text{y}-6={\frac{-14}{23}}(\text{x}+7)$
$23\text{y}-138=-14\text{x}-98$
$14\text{x}+23\text{y}-40=0$
View full question & answer→Question 1075 Marks
Find the equations to the altitudes of the triangle whose angular points are A(2, -2), B(1, 1) and C(-1, 0).
AnswerAD,BE and CF are the three altitudes of the triangle We know, Slope of AD × Slope of BC = -1; AD passes through A(2, -2) Slope of BE × Slope of AC = -1; AD passes through A(1, 1) Slope of CF × Slope of AB = -1; AD passes through C(-1, 0) Slope of $\text{BC}=\frac{0-1}{-1-1}=\frac{-1}{-2}=\frac{1}{2} \Rightarrow$ Slope of AD = -2 Slope of $\text{AC}=\frac{0-(-2)}{-1-2}=\frac{2}{-3}=\frac{-2}{3}\Rightarrow$ Slope of $\text{BE}=\frac{3}{2}$ Slope of $\text{AB}=\frac{1+2}{1-2}=\frac{3}{-1}=-3\Rightarrow$ Slope of $\text{CF}=\frac{1}{3}$ So, for AD, we have $\text{y}-\text{y}_1=\text{m}(\text{x}-\text{x})_1$$\Rightarrow\ \text{y}-(-2)=-2(\text{x}-2)$
$\Rightarrow\ \text{y}+2=-2\text{x}+4$
$\Rightarrow\ \text{2x}+\text{y}-2=0$
And, for BE, WE have
$\Rightarrow\text{y}-\text{y}_1=\text{m}(\text{x}-\text{x}_1)$
$\Rightarrow\text{y}-1=\frac{3}{2}(\text{x}-1)$
$\Rightarrow\ \text{2y}-\text{3x}+1=0$
And, for CF, We have
$\text{y}-\text{y}=\text{m}(\text{x}-\text{x}_1)$$\Rightarrow\ \text{y}-0=\frac{1}{3}(\text{x}+1)$
$\Rightarrow\ \text{x}-\text{3y}+1=0$
View full question & answer→Question 1085 Marks
Two sides of an isosceles triangle are given by the equations $7x - y + 3 = 0$ and $x + y - 3 = 0$ and its third side passes through the point $(1, -10)$. Determine the equation of the third side.
AnswerSolving $7x - y + 3 = 0$ and $x + y - 3 = 0$ we get, $A (0, 3)$
The slope of $7x - y + 3 = 0 (m_1)$ and $x + y - 3 = 0 (m_2)$ are $7$ and $-1$ respectively.
Any line through the point $(1, -10)$ is
$y + 10 = m (x - 1) ...(i)$
Since it make equal angle say $\theta$ with the given lines, therefore
$\tan\theta=\frac{\text{m}-7}{1+7\text{m}}=\frac{\text{m}-(-1)}{1+\text{m}(-1)}$
$\Rightarrow\text{m}=-3\text{ or }\frac{1}{3}$
Putting in $(i)$
$y + 10 = -3 (x - 1)$
$y + 10 = -3x + 3$
$3x + y + 7 = 0$
$\text{y}+10=\frac{1}{3}(\text{x}-1)\Rightarrow\frac{\text{x}}{3}-\frac{1}{3}$
$3\text{y}-\text{x}+31=0$
View full question & answer→Question 1095 Marks
Prove that the area of the parallelogram formed by the lines 3x - 4y + a = 0, 3x - 4y + 3a= 0, 4x - 3y - a = 0 and 4x - 3y - 2a = 0 is $\frac{2\text{a}^2}{7}$ sq.units.
AnswerThe area of a parallelogram is
$=\frac{|\text{c}_1-\text{d}_1||\text{c}_2-\text{d}_2|}{|\text{a}_2\text{b}_1-\text{b}_2\text{a}_1|}$
$=\frac{|-\text{a}+2\text{a}||3\text{a}-\text{a}|}{|3(-3)-4(-4)|}$
$=\frac{\text{a}\times2\text{a}}{7}$
$=\frac{2}{7}\text{a}^2$
Hence proved.
View full question & answer→Question 1105 Marks
Find the projection of the point $(1, 0)$ on the line joining the points $(-1, 2)$ and $(5, 4).$
AnswerLet $AB$ be the line, $A = (-1, 2), B = (5, -4)$
Then, equation of line AB is
$\text{y}-\text{y}_1=\text{m}(\text{x}-\text{x}_1)$
$\text{y}-\text{y}_1=\frac{\text{y}_2-\text{y}_1}{\text{x}_2-\text{x}_1}(\text{x}-\text{x}_1)$
$\text{y}-2=\frac{4-2}{5+1}(\text{x}+1)$
$\text{y}-2=\frac{2}{6}(\text{x}+1)$
$3\text{y}-\text{x}=7 \ ...(1)$
Slope $=\frac{1}{3}.$
Let P point $(1, 0)$ be the given point
Let $Q(x_1, y_1)$ be the projection of P
Slope of PQ = -3 $\big[\text{PQ}\perp\text{AB},\text{m}_1\text{m}_2=-1\big]$
Eq of PQ,
$\text{y} - 0 = -3(\text{x} - 1)$
$\text{y} = -3\text{x} + 3 \ ...(2)$
Solving (1) and (2)
$3\text{y}-\Big(\frac{\text{y}-3}{-3}\Big)=7$
$-9\text{y}-\text{y}+3=-21$
$-10\text{y}=-24$
$\Rightarrow\frac{12}{5}=-3\text{x}+3$
$-3\text{x}=+\frac{12}{5}-3=\frac{+12-15}{5}=\frac{-3}{5}$
$\text{x}=\frac{1}{5}$
$\therefore\text{N}\Big(\frac{1}{5},\frac{`12}{5}\Big)$
View full question & answer→Question 1115 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 Q(5, 2) be the mirror image of P(2, -1) with respect to the line mirror AB × (ax + by + c = 0)
Then,
(Slope of AB) × (Slope of PQ) = -1
$\frac{-\text{a}}{\text{b}}\times\Big(\frac{2-1}{5-2}\Big)=-1$
$\frac{-\text{a}}{\text{b}}\times\frac{1}{3}=-1$
$-\text{a}=-3\text{b}$
$\text{a}=3\text{b} \ ...(1)$
and
(R) mid point of PQ should line in AB, as PQ perpendicularlly bisects AB.
$\therefore$ Coordinates of R are $\Big(\frac{5+2}{2},\frac{2+1}{2}\Big)=\Big(\frac{7}{2},\frac{3}{2}\Big)$
$\therefore\frac{7}{2}\text{a}+\frac{3}{2}\text{b}+\text{c}=0$
$7\text{a}+3\Big(\frac{\text{a}}{3}\Big)+2\text{c}=0 \ \Big[\because\text{b}=\frac{\text{a}}{3} \ \text{from(1)}\Big]$
$8\text{a}+2\text{c}=0$
or, $-4\text{a}=6 \ ...(2)$
$\therefore$ equation of line is ax + by + c = 0
or, $\text{ax}+\frac{\text{a}}{3}\text{y}-4\text{a}=0$
or, $3\text{x}+\text{y}-12=0$
View full question & answer→Question 1125 Marks
If $\theta$ is the angle which the straight line joining the points $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ subtends at the origin, prove that $\tan\theta=\frac{\text{x}_2\text{y}_1-\text{x}_1\text{y}_2}{\text{x}_1\text{x}_2+\text{y}_1\text{y}_2}$ and $\cos\theta=\frac{\text{x}_1\text{x}_2+\text{y}_1\text{y}_2}{\sqrt{\text{x}_1^2+\text{y}_1^2}\sqrt{\text{x}_2^2+\text{y}_2^2}}$
AnswerLet $l_1$, be the line joining $AO$ and
Let $l_2$, be the line joining $BO$
Then, line $l_1$ is $\text{y}-0=\Big(\frac{0-\text{x}_1}{0-\text{y}_1}\Big)(\text{x}-0)$
$\text{y}\text{y}_1=\text{x}_1\text{x}=0$
Then, $\text{m}_1=\frac{\text{x}_1}{\text{y}_1}$
Then line $l_2$ is $\text{y}-0=\Big(\frac{0-\text{x}_2}{0-\text{y}_2}\Big)(\text{x}-0)$
Then, $\text{m}_2=\frac{\text{x}_2}{\text{y}_2}$
Then, $\Rightarrow\tan\theta=\Big|\frac{\text{m}_1-\text{m}_2}{1+\text{m}_1\text{m}_2}\Big|=\Bigg|\frac{\frac{\text{x}_1}{\text{y}_1}-\frac{\text{x}_2}{\text{y}_2}}{1+\frac{\text{x}_1}{\text{y}_1}\frac{\text{x}_2}{\text{y}_2}}\Bigg|$
$=\Big|\frac{\text{x}_1\text{y}_2-\text{y}_1\text{x}_2}{\text{y}_1\text{y}_2+\text{x}_1\text{x}_2}\Big|$
From triangle,
$\text{AC}=\sqrt{(\text{AB})^2+(\text{BC})^2}$
$=\sqrt{(\text{m}_1^2+\text{m}_2^2-2\text{m}_1\text{m}_2)+(1+\text{m}_1\text{m}_2)^2}$
$=\sqrt{\text{m}_1^2+\text{m}_2^2-2\text{m}_1\text{m}_2+1+\text{m}_1^2\text{m}_2^2+2\text{m}_1\text{m}_2}$
$=\sqrt{\text{m}_1^2+\text{m}_2^2+1+\text{m}_1^2\text{m}_2^2}$
$\cos\theta=\frac{\text{BC}}{{\text{AC}}}=\frac{1+\text{m}_1\text{m}_2}{\sqrt{\text{m}_1^2+\text{m}_2^2+\text{m}_1^2\text{m}_2^2+1}}$
$=\frac{1+\frac{\text{x}_1}{\text{y}_1}\frac{\text{x}_2}{\text{y}_2}}{\sqrt{\frac{\text{x}_1^2}{\text{y}_1^2}+\frac{\text{x}_2^2}{\text{y}_2^2}+\frac{\text{x}_1^2\text{x}_2^2}{\text{y}_1^2\text{y}_2^2}}+1}$
$=\frac{\frac{\text{y}_1\text{y}_2+\text{x}_1\text{x}_2}{\text{y}_1\text{y}_2}}{\sqrt{\frac{\text{x}_1^2\text{y}_2^2+\text{x}_2^2\text{y}_1^2+\text{x}_1^2\text{x}_2^2+\text{y}_1^2\text{y}_2^2}{\text{y}_1^2\text{y}_2^2}}}$
$=\frac{\text{y}_1\text{y}_2+\text{x}_1\text{x}_2}{\sqrt{\text{x}_1^2(\text{y}_2^2+\text{x}_2^2)+\text{y}_1^2(\text{y}_2^2+\text{x}_2^2)}}$
$=\frac{\text{y}_1\text{y}_2+\text{x}_1\text{x}_2}{\sqrt{\text{x}_1^2+\text{y}_1^2}\sqrt{\text{y}_2^2+\text{x}_2^2}}$
Hence proved.
View full question & answer→Question 1135 Marks
Find the equation of the straight line drawn through the point of intersection of the lines x + y = 4 and 2x - 3y = 1 and perpendicular to the line cutting off intercepts 5, 6 on the axes.
AnswerThe required line is
$(\text{x}+\text{y}-4)+\lambda(2\text{x}-3\text{y}-1)=0$
or, $\text{x}(1+2\lambda)+\text{y}(1-3\lambda)-4-\lambda=0$
And it is perpendicular to $\frac{\text{x}}{5}+\frac{\text{y}}{6}=1$
$\therefore(\text{slope of required line})\times\Big(\text{slope of} \ \frac{\text{x}}{5}+\frac{\text{y}}{6}=1\Big)=-1$
$\Rightarrow-\Big(\frac{1+2\lambda}{1-3\lambda}\Big)\times\frac{-6}{5}=-1$
$\Rightarrow\frac{1+2\lambda}{1-3\lambda}=\frac{-5}{6}$
$\Rightarrow6+12\lambda=-5+15\lambda$
$\Rightarrow11=3\lambda$ or $\lambda=\frac{11}{3}$
$\therefore$ The required line is
$(\text{x}+\text{y}-4)+\frac{11}{3}(2\text{x}-3\text{y}-1)=0$
$3\text{x}+3\text{y}-12+22\text{x}-33\text{y}-11=0$
$25\text{x}-30\text{y}-23=0$
View full question & answer→Question 1145 Marks
Prove that the following sets of three lines are concurrent:
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1, \frac{\text{x}}{\text{b}}+\frac{\text{y}}{\text{a}}=1$ and $\text{y}=\text{x}.$
Answer$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1, \frac{\text{x}}{\text{b}}+\frac{\text{y}}{\text{a}}=1$ and $\text{y}=\text{x}$
bx + ax = ab, ax + by = ab
put y = x
bx + ax = ab, ax + bx = ab
Hence the lines are concurrent
View full question & answer→Question 1155 Marks
Prove that the area of the parallelogram formed by the lines $a_1x + b_1y + c_1 = 0, a_1x + b_1y+ d_1 = 0, a_2x + b_2y + c_2 = 0, a_2x + b_2y + d_2 = 0$ is $\Big|\frac{(\text{d}_1-\text{c}_1)(\text{d}_2-\text{c}_2)}{\text{a}_1\text{b}_2-\text{a}_2\text{b}_1}\Big|$ sq.units.
Deduce the condition for these lines to form a rhombus.
AnswerLet $ABCD$ be a parallelogram the equation of whose sides $AB, BC, CD$ and $DA$ are $a_1x + b_1y + c_1 = 0, a_1x + b_1y+ d_1 = 0, a_2x + b_2y + c_2 = 0, a_2x + b_2y + d_2 = 0.$
Let $p_1$ and $p_2$ be the distance between the pairs of parallel side of $ABCD.$
$\sin\theta\frac{\text{p}_1}{\text{AD}}=\frac{\text{p}_2}{\text{AB}}$
$\therefore\text{AD}=\frac{\text{p}_1}{\sin\theta}$ and $\text{AB}=\frac{\text{p}_2}{\sin\theta}$
Area of ABCD $=\text{AB}\times\text{p}_1=\frac{\text{p}_1\text{p}_2}{\sin\theta}$
or $\Rightarrow\text{AD}\times\text{p}_2=\frac{\text{p}_1\text{p}_2}{\sin\theta}$
Now,
$m_1 =$ slope of $\text{AB}=\frac{\text{a}_1}{\text{b}_1}$
$m_2 =$ slope of $\text{AD}=\frac{-\text{a}_2}{\text{b}_1}$
Since $\theta$ is angle between AB and AC.
$\tan\theta=\frac{\text{m}_1-\text{m}_2}{1+\text{m}_1\text{m}_2}$
$=\frac{\frac{-\text{a}_2}{\text{b}_2}+\frac{\text{a}_1}{\text{b}_1}}{1-\frac{\text{a}_1\text{a}_2}{\text{b}_1\text{b}_2}}$
$\tan\theta=\frac{\text{a}_2\text{b}_1-\text{a}_1\text{b}_2}{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2}\Rightarrow\sin\theta=\frac{\text{a}_2\text{b}_1-\text{a}_1+\text{b}_2}{\sqrt{\big(\text{a}_1^2+\text{b}_1^2\big)\big(\text{a}_2^2+\text{b}_2^2\big)}}$
$p_1 =$ Distance between AB and AD.
$=\Bigg|\frac{\text{c}_1-\text{d}_1}{\sqrt{\text{a}_1^2+\text{b}_1^2}}\Bigg|$
$p_2 =$ Distance between AD and BC.
$=\Bigg|\frac{\text{c}_2-\text{d}_2}{\sqrt{\text{a}_2^2+\text{b}_2^2}}\Bigg|$
$\therefore$ Area of parallelogram is
$\frac{|\text{c}_1-\text{d}_1||\text{c}_2-\text{d}_2|}{|\text{a}_2\text{b}_1-\text{a}_1\text{b}_2|}$
$\Big|\frac{(\text{d}_1-\text{c}_1)(\text{d}_2-\text{c}_2)}{\text{a}_1\text{b}_2-\text{a}_2\text{b}_1}\Big|$
Hence, proved.
Rhombus is a paralleogram with all side equal.
$\therefore$ $p_1 = p_2$
$\therefore$ Modifing the formula of area of parallelogram divided above.
The area of rhombus
$=\frac{\text{p}_1\text{p}_2}{\sin\theta}$
$=\frac{2\text{p}_1}{\sin\theta}=\frac{2\text{p}_2}{\sin\theta}$
$=2\Big|\frac{(\text{c}_1-\text{d}_1)}{\text{a}_1\text{b}_1-\text{a}_1\text{b }_2}\Big|$ or $2\Big|\frac{(\text{c}_2-\text{d}_2)}{\text{a}_2\text{b}_1-\text{b}_2\text{a}_1}\Big|$
View full question & answer→Question 1165 Marks
Prove that the following sets of three lines are concurrent:
3x - 5y - 11 = 0, 5x + 3y - 7 = 0 and x + 2y = 0
Answer3x - 5y - 11 = 0, 5x + 3y - 7 = 0 and x + 2y = 0
3x - 5y - 11 = 0 ...(1)
5x + 3y - 7 = 0 ...(2)
x + 2y = 0 ...(3)
Solving (1) and (2)
x = -2y
5(-2y) + 3y - 7 = 0
-10y + 3y - 7 = 0
-7y = y
y = -1
⇒ x = 2
substituting x and y in (1)
3(2) - 5(-1) - 11 = 0
6 + 5 - 11 = 0
0 = 0
Hence, the lines are concurrent
View full question & answer→Question 1175 Marks
Find the equations of two straight lines passing through $(1, 2)$ and making an angle of $60^\circ $ with the line $x + y = 0$. Find also the area of the triangle formed by the three lines.
Answer$A C$ and $B C$ are inclided to $(A B) x+y=0$ at an angle of $60^{\circ}$
$\therefore \triangle ABC$ is equilateral triangle.
The slope of $A B$ is -1 and let slope of $A C$ be $m_1$ $\tan 60^{\circ}=\frac{ m _1+1}{1- m _1}$ or $\sqrt{3}\left(1- m _1\right)= m _1+1$
$\sqrt{3}-1= m _1+\sqrt{3} m$
$\Rightarrow m_1=\frac{\sqrt{3}-1}{\sqrt{3}+1}=2-\sqrt{3}$
and, slope of $B C$ is $m_2$
$\tan60^\circ=\frac{\text{m}_2-1}{1+\text{m}_1}=\sqrt3$
$\therefore\text{m}_2=\sqrt3+2$
$\therefore$ Equation of $A C$ and $B C$ are
$\text{y}-2=(2-\sqrt{3})(\text{x}-1) \ ...(\text{i})$
using (i) and $x+y=0$
A is $\Big(\frac{-1-\sqrt3}{2},\frac{1+\sqrt3}{2}\Big)$A
$AC$ is $\sqrt{\Big(\frac{2+1+\sqrt{3}}{2}\Big)^2+\Big(\frac{3-\sqrt{3}}{2}\Big)^2}$
$=\sqrt\frac{9+3+6\sqrt{3}+9+3-6\sqrt{3}}{4}$
$\text{AC}=\sqrt{\frac{24}{6}}$
$=\sqrt6$
The area of $\Delta\text{ABC}$
$=\frac{\sqrt3}{4}(\text{AC})^2$
$=\frac{\sqrt3}{4}\times(\sqrt6)^2$
$=\frac{3}{2}\sqrt{3} \text{ sq untis.}$
View full question & answer→Question 1185 Marks
Find the image of the point (3, 8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror.
AnswerLet the image of p(3, 8) in x + 3y = 7 be $\text{Q}(\alpha,\beta)$
Then.
PQ is perpendicular bisected at R.
Then,
$\text{R}=\Big(\frac{\alpha+3}{2},\frac{\beta+8}{2}\Big)$
and lie on x + 3y = 7
$\frac{\alpha+3}{2}+\frac{3\beta+24}{2}=7$
$\alpha+3+3\beta+24=14$
$\alpha+3\beta=-13 \ ...(1)$
And since PQ is perpendicular to
x + 3y = 7
(Slope of line) × (Slope of PQ) = -1
$\frac{-1}{3}\times\frac{\beta-8}{\alpha-3}=-1$
$\beta-8=3\alpha-9$
$\beta-3\alpha=-1 \ ...(2)$
Solving (1) and (2)
$\beta=-4, \alpha=-1$
$\therefore$ Q is (-1, -4)
View full question & answer→Question 1195 Marks
Find the equations to the sides of an isosceles right angled triangle the equation of whose hypotenues is $3x + 4y = 4$ and the opposite vertex is the point $(2, 2).$
AnswerLet the isosceles right triangle be.
$AC = 3x + 4y = 4$
$C(2, 2)$
Then, slope of $\text{AC}=\frac{-3}{4}$
$AB = BC \big[\therefore$ It is an isoscales right triangle $\big]$
Then, angle between $(AB$ and $AC)$ and $(BC$ and $AC)$ is $45^\circ .$
$\tan\frac{\pi}{4}=\frac{\text{m}_1-\big(\frac{-3}{4}\big)}{1+\big(\frac{-3}{4}\big)\text{m}_1}$ [when m_1 = slope of BC]
$1=\frac{\text{m}_1+\frac{3}{4}}{1-\frac{3}{4}\text{m}}$
$4-3\text{m}_1=4\text{m}_1+3$
$7\text{m}_1=1$
$\text{m}_1=\frac{1}{7}$
and, $\text{AB}\perp\text{BC}$
$\therefore ($Slope of $AB) \times ($slope of $BC) = -1$
$\text{m}_2\times\frac{1}{7}=-1$
$\text{m}_2=-7$
The equation of $BC$ is
$(\text{y}-2)=\frac{1}{7}(\text{x}-2)$
$7\text{y}-14=\text{x}-2$
$\text{x}-7\text{y}+12=0$
and
The equation of $AB$ is
$(\text{y}-2)=-7(\text{x}-2)$
$\text{y}-2=-\text{x}+14$
$\text{y}+7\text{x}-16=0$
View full question & answer→Question 1205 Marks
Find the coordinates of the vertices of a triangle, the equations of whose sides are:
$x + y - 4 = 0, 2x - y + 3 = 0$ and $x - 3y + 2 = 0$
Answer
- The point of in intersection of two side will give the vertex
$x + y - 4 = 0 ...(1)$
$2x - y + 3 =0 ...(2)$
$x - 3y + 2 = 0 ...(3)$
solving $(1)$ and $(2)$
$x + y = 4$
$y = 4 - x$
Putting $y$ in $(2)$
$2x - (4 - x) + 3 = 0$
$2x - 4 + x + 3 = 0$
$3x - 1 = 0$
$\text{x}=\frac{1}{3}$
Putting $x$ in $(1)$
$\frac{1}{3}+\text{y}-4=0$
$\text{y}=4-\frac{1}{3}=\frac{11}{3}$
$\therefore$ one vertex is $\bigg(\frac{1}{3},\frac{11}{3}\bigg)$
- Solving $(2)$ and $(3),$ we get
$y = 2x + 3$ and putting in $(3)$
$x - 3y + 2 = 0$
$x - 3(2x + 3) + 2 = 0$
$x - 6x - 9 + 2 = 0$
$-5x = +7$
$\text{x}=\frac{-7}{5}$
$\Rightarrow \text{y}=2\text{x}+3=2\Big(\frac{-7}{5}\Big)+3=\frac{-14}{5}+3=\frac{1}{5}$
- $\therefore$ Second vertex is $\Big(\frac{-7}{5},\frac{1}{5}\Big)$
For find vertex
$x + y - 4 = 0$
$x - 3y + 2 = 0$
$x = 4 - y$
$4 - y - 3y + 2 = 0$
$4 - 4y -3y + 2 = 0$
$-4y = -6$
$\text{y}=\frac{3}{2}$
$\Rightarrow\text{x}=4-\text{y}$
$4-\frac{3}{2}$
$\frac{8-3}{2}=\frac{5}{2}$
$\therefore$ The third vertex is $\Big(\frac{5}{2},\frac{3}{2}\Big)$ View full question & answer→Question 1215 Marks
The equation of the base of an equilateral triangle is x + y = 2 and its vertex is (2, -1). Find the length and equations of its sides.
AnswerThe slope of AB = -1
Let slope of AC be m
Then,
$\tan60^\circ=\frac{\text{m+1}}{1-\text{m}}$
$\text{m}=2-\sqrt3$
And similarly slope of $\text{AB}=2+\sqrt3.$
Equation of AC and AB are
$(\text{y+1})=(2-\sqrt3)(\text{x}-2)$
or, $(2-\sqrt3)\text{ x}-\text{y}-5+2\sqrt3=0 \ ...(\text{i})$
and,
$(\text{y}-1)=(2+\sqrt3)(\text{x}-2)$
or, $(2+\sqrt3)\text{ x}-\text{y}-5-2\sqrt{3}=0 \ ...(\text{ii})$
On solving (i) with x + y = 2, we get
$\text{A}\Big(\frac{21-11\sqrt3}{6},\frac{11\sqrt3-9}{6}\Big)$
$\text{AB}=\text{AC}=\text{BC}$
$=\sqrt{\Big(\frac{21-11\sqrt3-1}{6}\Big)^2+\Big(\frac{11\sqrt3-9-1}{6}\Big)^2}$
$=\sqrt{\frac{225+363-330\sqrt{3}+363+225-330\sqrt{3}}{36}}$
$=\sqrt{\frac{2}{3}}$
View full question & answer→Question 1225 Marks
For what value of λ are the three lines 2x - 5y + 3 = 0, 5x - 9y + λ = 0 and x − 2y + 1 = 0 concurrent?
AnswerThe three lines are concurrent if they have the common point of intersection.
2x - 5y + 3 = 0 ...(1)
x - 2y + 1 = 0 ...(2)
Solving (1) and (2)
2x = 5y - 3
$\text{x}=\frac{5\text{y}-3}{2}$
$\frac{5\text{y}-3}{2}-2\text{y}+1=0$
5y - 3 - 4y + 2 = 0
y = 0
$\Rightarrow\text{x}=\frac{5\text{y}-3}{2}=\frac{5-3}{2}=\frac{2}{2}=1$
Substituting x and y is 5x - 9y + λ = 0
5(1) - 9(1) + λ = 0
5 - 9 + λ = 0
λ = 4
View full question & answer→Question 1235 Marks
Find the values of $\alpha$ so that the point $\text{p}(\alpha^2,\alpha)$ lies inside or on the triangle formed by the lines $x - 5y + 6 = 0, x - 3y + 2 = 0$ and $x - 2y - 3 = 0.$
AnswerLet $\text{ABC}$ be the triangle of the equations whose sides $\text{AB, BC}$ and $\text{CA}$ are respectively $x - 5y + 6 = 0, x - 3y + 2 = 0$ and $x - 2y - 3 = 0$
The coordinates of the vertices are $A(9, 3), B(4, 2)$ and $C(13, 5).$
If the point $\text{p}(\alpha^2,\alpha)$ lies $n$ side the $\triangle\text{ABC},$ then
- $A$ and $P$ must be on the same side of $\text{BC.}$
- $B$ and $P$ must be on the same side of $\text{AC.}$
- $C$ and $P$ must be on the same side of $\text{AB.}$
Now,
$A$ and $P$ must be on the same side of $BC$ if,
$\big(9(1)+3(-3)+2\big)\Big(\alpha^2-3\alpha+2\Big)>0$
$(9-9+2)\big(\alpha^2-3\alpha+2\big)>0$
$\alpha^2-3\alpha+2>0$
$(\alpha-1)(\alpha-2)>0 $
$\alpha\in(-\infty,1)\cup(2,\infty) \ ...(\text{i})$
$B$ and $P$ must be on the same side of $AC$ if,
$\big(13(1)+5(-5)+6\big)\Big(\alpha^2-5\alpha+6\Big)>0$
$\Rightarrow(-6)\big(\alpha^2-5\alpha+6\big)>0$
$\Rightarrow\alpha^2-5\alpha+6<0$
$\Rightarrow(\alpha-2)(\alpha-3)<0 $
$\Rightarrow\alpha\in(2, 3) \ ...(\text{ii})$
$C$ and $P$ must be on the same side of $AB$ if,
$\big(4(1)+2(-2)-3\big)\Big(\alpha^2-2\alpha-3\Big)>0$
$(-3)\big(\alpha^2-2\alpha-3\big)>0$
$\alpha^2-2\alpha-3<0$
$(\alpha-3)(\alpha+1)<0 $
$\Rightarrow\alpha\in(-1, 3) \ ...(\text{iii})$
From $i, ii$, and $iii$
$\alpha\in[2, 3]$ View full question & answer→Question 1245 Marks
Show that the straight lines given by $(2 + k)x + (1 + k)y = 5 + 7k$ for different values of $k$ pass through a fixed point. Also, find that point.
Answer$(2 + k)x + (1 + k)y = 5 + 7k$
or, $(2x + y - 5) + k(x + y -7) = 0$
It is of the form $L_1 + KL_2 = 0$ i.e., the equation of line passing through the intersection of $2$ lines $L_1 $ and $L_2.$
So, it represents a line passing through $2x + y - 5 = 0$ and $x + y - 7 = 0.$
Solving the two equation we get, $(-2, 9).$ Which is the fixed point through which the given line pass. For any value of $k.$
View full question & answer→