MCQ 512 Marks
If a and b are two arbitrary constants, then the straight line (a-2b)x + (a+3b)y + 3a+ 4b = 0 will pass through
Answer(A)
$(a-2 b) x+(a+3 b) y+3 a+4 b=0$ or $a (x+y+3)+ b (-2 x+3 y+4)=0$, which represents a family of straight lines through point of intersection of $x+y+3=0$ and $-2 x+3 y+4=0$
i e, $(-1,-2)$.
View full question & answer→MCQ 522 Marks
Two sides of a rhombus are along the lines, x - y + 1 = 0 and 7x - y - 5 = 0 If its diagonals intersect at (-1,-2), then which one of the following is a vertex of this rhombus?
- A
- B
- C
$\left(\frac{1}{3},-\frac{8}{3}\right)$
- D
$\left(-\frac{10}{3},-\frac{7}{3}\right)$
View full question & answer→MCQ 532 Marks
Two consecutive sides of a parallelogram are 4x + 5y = 0 and 7x + 2y = 0 If the equation to one diagonal is 11x + 7y = 9, then the equation of the other diagonal is
MCQ 542 Marks
The diagonal passing through origin of a quadrilateral formed by x = 0 y = 0 x + y = 1
and 6x + y = 3 is
MCQ 552 Marks
Equations of diagonals of square formed by lines x = 0, y = 0, x = 1 and y = 1 are
- A
- B
- C
$2 y=x, y+x=\frac{1}{3}$
- D
View full question & answer→MCQ 562 Marks
Equations of lines which passes through the points of intersection of the lines 4x - 3y - 1 = 0 and 2x - 5y + 3 = 0 and are equally inclined to the axes are
- A
$y \pm x=0$
- B
$y-1= \pm 1(x-1)$
- C
$x-1= \pm 2(y-1)$
- D
$y \pm x=2$
View full question & answer→MCQ 572 Marks
The equation of the straight line passing through the point of intersection of 3x + 2y + 5 = 0,5x - 6y - 1 = 0 and perpendicular to the line 3x - 5y + 11 = 0 is
Answer(C)
$5 x-6 y-1=0$ ...(i)
$3 x+2 y+5=0$ ...(ii)
On solving (i) and (ii), we get $x=-1, y=-1$ Slope of line $3 x-5 y+11=0$ is $\frac{3}{5}$.
Slope of line $3 x-5 y+11=0$ is $\frac{3}{5}$.
Slope of line perpendicular to above line $=\frac{-5}{3}$
∴ Equation of line passing through $(-1,-1)$ and having slope $-\frac{5}{3}$ is
$(y+1)=-\frac{5}{3}(x+1)$
$\Rightarrow 3 y+3=-5 x-5$
$\Rightarrow 5 x+3 y+8=0$
View full question & answer→MCQ 582 Marks
The line passing through the point of intersection of x + y = 2, x - y = 0 and is parallel
to x + 2y = 5 is
Answer(D)
Required equation of line which is parallel to
$x+2 y=5 \text { is } x+2 y+k=0$ ...(i)
Given equation of lines are
$x+y=2$ ...(i)
$x-y=0$ ...(ii)
Adding (ii) and (iii), we get $2 x=2 \Rightarrow x=1$
From (iii), we get $y=1$
∴ Point of intersection is $(1,1)$.
Putting $x=1, y=1$ in (i),
we get $k =-3$
∴ the required equation of line is $x+2 y=3$.
View full question & answer→MCQ 592 Marks
For the straight lines given by the equation $(2+ k ) x+(1+ k ) y=5+7 k$, for different values of k which of the following statements is true?
- A
- ✓
Lines pass through the point (-2, 9)
- C
Lines pass through the point (2,-9)
- D
AnswerCorrect option: B. Lines pass through the point (-2, 9)
(B)
Putting $k =1,2$, we get
$3 x+2 y=12$ ...(i)
$4 x+3 y=19$ ...(ii)
The given lines are not parallel.
Hence on solving them, we get
$x=-2, y=9$
Therefore, the lines pass through $(-2,9)$
View full question & answer→MCQ 602 Marks
If the point of intersection of the lines 2ax + 4ay + c = 0 and 7bx + 3by - d = 0 lies in the 4th quadrant and is the two axes, where a, b, c and d are non-zero numbers, then ad: bc equals to
Answer(B)
Let $(\alpha,-\alpha)$ be the point of intersection.
Then, $2 a \alpha-4 a \alpha+ c =0$
$\Rightarrow \alpha=\frac{ c }{2 a }$ ...(i)
and $7 b \alpha-3 b \alpha- d =0$
$\Rightarrow \alpha=\frac{ d }{4 b}$ ...(ii)
From (i) and (ii), we get
$\frac{c}{2 a}=\frac{d}{4 b}$
$\Rightarrow \frac{ ad }{ bc }=2$
$\Rightarrow ad : bc =2: 1$
View full question & answer→MCQ 612 Marks
The point of intersection of the lines $\frac{x}{a}+\frac{y}{b}=1$ and $\frac{x}{b}+\frac{y}{a}=1$ lies on the line
Answer(D)
Intersection point of the line is $\left(\frac{a b}{a+b}, \frac{a b}{a+b}\right)$, which is satisfying all the equations given in options (A), (B) and (C).
Hence, (D) is correct.
View full question & answer→MCQ 622 Marks
If $a+b+c=0$ and $p \neq 0$, the lines $ax +( b + c ) y= p , \quad bx +( c + a ) y= p$ and $c x+(a+b) y=p$
Answer(A)
By the given condition of $a+b+c=0$, the three lines reduce to
$x-y=\frac{p}{a} \text { or } \frac{p}{b} \text { or } \frac{p}{c}(p \neq 0) .$
All these lines are parallel. Hence, they do not intersect in finite plane.
View full question & answer→MCQ 632 Marks
If the lines ax y + 1 = 0, x + by + 1 = 0 and x + y + c = 0 (a, b, c being distinct and different from 1) are concurrent, then $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=$
Answer(B)
If the given lines are concurrent, then
$\left|\begin{array}{lll}a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c\end{array}\right|=0 \Rightarrow\left|\begin{array}{ccc}a & 1-a & 1-a \\ 1 & b-1 & 0 \\ 1 & 0 & c-1\end{array}\right|=0$
$\ldots\left[\right.$ By $C _2 \rightarrow C _2- C _1$ and $\left.C _3 \rightarrow C _3- C _1\right]$
$\Rightarrow a ( b -1)( c -1)-( b -1)(1- a )$$-(c-1)(1-a)=0$
$-(c-1)(1-a)=0$
$\Rightarrow \frac{ a }{1- a }+\frac{1}{1- b }+\frac{1}{1- c }=0$
.... [Divide by $(1- a )(1- b )(1- c )$ ]
$\Rightarrow-\frac{(1-a-1)}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=0$
$\Rightarrow \frac{-(1-a)}{1-a}+\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=0$
$\Rightarrow-1+\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=0$
$\Rightarrow \frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=1$
View full question & answer→MCQ 642 Marks
The straight lines x + 2y - 9 = 0 3x + 5y - 5 = 0 and ax + by - 10 are concurrent, if the straight line 35x - 22y + 1 = 0 passes through the point
Answer(A)
The three lines are concurrent, if
$\left|\begin{array}{ccc}1 & 2 & -9 \\ 3 & 5 & -5 \\ a & b & -1\end{array}\right|=0$
$\Rightarrow 35 a -22 b+1=0$
which is true if the line $35 x-22 y+1=0$ passes through $( a , b )$.
View full question & answer→MCQ 652 Marks
The straight line passing through the point of intersection of the straight lines x + 2y - 10 = 0 and 2x + y + 5 = 0 is
Answer(B)
$\left|\begin{array}{ccc}1 & 2 & -10 \\2 & 1 & 5 \\5 & 4 & 0\end{array}\right|=1(0-202(-25)-10(3)=0$
Hence, option (B) is the correct answer.
View full question & answer→MCQ 662 Marks
The value of $\lambda$ for which the lines 3x + 4y = 5, 5x + 4y = 4 and $\lambda x+4 y=6$ meet at a point is
Answer(B)
Given lines are concurrent.
∴$\begin{array}{l}\left|\begin{array}{lll}3 & 4 & -5 \\5 & 4 & -4 \\\lambda & 4 & -6\end{array}\right| \\\Rightarrow 3(-24+16)-4(-30+4 \lambda)-5(20-4 \lambda)=0 \\\Rightarrow \lambda=1\end{array}$
View full question & answer→MCQ 672 Marks
The lines
(p - q) x + (q - r) y + (r - p) = 0
(q - r) x + (r - p) y + (p - q) = 0
(r - p) x + (p - q) y + (q - r) = 0 are
Answer(C)
$\left|\begin{array}{lll}p-q & q-r & r-p \\ q-r & r-p & p-q \\ r-p & p-q & q-r\end{array}\right|=\left|\begin{array}{lll}0 & q-r & r-p \\ 0 & r-p & p-q \\ 0 & p-q & q-r\end{array}\right|=0$
$\ldots\left[ By C _1 \rightarrow C _1+ C _2+ C _3\right]$
Hence, the lines are concurrent.
View full question & answer→MCQ 682 Marks
The lines ax + by + c = 0 where 3a + 2b + 4c = 0 are concurrent at the point
Answer(D)
Dividing both sides of relation $3 a+2 b+4 c=0$ by 4 , we get $\frac{3}{4} a+\frac{1}{2} b+c=0$, which shows that for all values of $a , b$ and c each member of the set of lines $a x+ b y+ c =0$ passes through the point $\left(\frac{3}{4}, \frac{1}{2}\right)$
View full question & answer→MCQ 692 Marks
The lines 2x + y - 1 = 0 ax 3y - 3 = 0 and 3x + 2y - 2 = 0 are concurrent for
- ✓
- B
- C
$-1 \leq a \leq 3$
- D
$a > 0$ only
Answer(A)
Given lines are concurrent, if $\left|\begin{array}{lll}2 & 1 & -1 \\ a & 3 & -3 \\ 3 & 2 & -2\end{array}\right|=0$
$\Rightarrow-\left|\begin{array}{lll}2 & 1 & 1 \\ a & 3 & 3 \\ 3 & 2 & 2\end{array}\right|=0$
This is true for all values of a because $C _2$ and $C _3$ are identical.
View full question & answer→MCQ 702 Marks
The opposite angular points of a square are (3, 4) and (1, -1). Then the co-ordinates of other two points are
- A
$D \left(\frac{1}{2}, \frac{9}{2}\right), B \left(-\frac{1}{2}, \frac{5}{2}\right)$
- B
$D \left(\frac{1}{2}, \frac{9}{2}\right), B \left(\frac{1}{2}, \frac{5}{2}\right)$
- C
$D \left(\frac{9}{2}, \frac{1}{2}\right), B \left(-\frac{1}{2}, \frac{5}{2}\right)$
- D
View full question & answer→MCQ 712 Marks
The equations of two equal sides of an isosceles triangle are 7x - y + 3 = 0 and x + y - 3 = 0 and the third side passes through the point (1, -10). The equation of the third side is
- A
$3 x-y-31=0$ or $x+y+7=0$
- B
$3 x-y+7=0$ or $x+3 y-31=0$
- ✓
$3 x+y+7=0$ or $x-3 y-31=0$
- D
Neither $3 x+y+7$ nor $x-3 y-31=0$
AnswerCorrect option: C. $3 x+y+7=0$ or $x-3 y-31=0$
(C)
Any line through $(1,-10)$ is given by
$y+10=m(x-1)$
Since, it makes equal angle say '$\alpha$' with the given lines $7 x-y+3=0$ and $x+y-3=0$
$\therefore \quad \tan \alpha=\frac{m-7}{1+7 m}=\frac{m-(-1)}{1+m(-1)}$
$\Rightarrow m =\frac{1}{3}$ or -3
Hence, the two possible equations of third side are $3 x+y+7=0, x-3 y-31=0$.
View full question & answer→MCQ 722 Marks
Equations of the two straight lines passing through the point (3, 2) and making an angle of \[45^{\circ}\] with the line x - 2y = 3 are
- A
$3 x+y+7=0$ and $x+3 y+9=0$
- ✓
$3 x-y-7=0$ and $x+3 y-9=0$
- C
$x+3 y-7=0$ and $x+3 y-9=0$
- D
AnswerCorrect option: B. $3 x-y-7=0$ and $x+3 y-9=0$
(B)
Slope of given line is $\frac{1}{2}$
Thus, $\tan 45^{\circ}= \pm \frac{m-\frac{1}{2}}{1+m \cdot \frac{1}{2}} \Rightarrow m=3$ or $\frac{-1}{3}$
Hence option (B) is correct.
View full question & answer→MCQ 732 Marks
A point on the straight line, 3x + 5y = 15 which is equidistant from the coordinate axes will lie
only in
- A
$4^{\text {th }}$ quadrant
- B
$1^{\text {st }}$ quadrant
- ✓
$1^{\text {st }}$ and $2^{\text {nd }}$ quadrants
- D
$1^{\text {st }}, 2^{\text {nd }}$ and $4^{\text {th }}$ quadrants
AnswerCorrect option: C. $1^{\text {st }}$ and $2^{\text {nd }}$ quadrants
(C)
A point on $3 x+5 y=15$ would be equidistant from the coordinate axes if it lies on $y=x$ or $y=-x$.
Solving $3 x+5 y=15$ and $y=x$,
we get $(x, y)=\left(\frac{15}{8}, \frac{15}{8}\right) \in 1^{\text {st }}$ quadrant
Solving $3 x+5 y=15$ and $y=-x$,we get
$(x, y)=\left(-\frac{15}{2}, \frac{15}{2}\right) \in 2^{\text {nd }}$ quadrant
View full question & answer→MCQ 742 Marks
If $m_1, m_2\left(m_1>m_2\right)$ are the slopes of the lines which make an angle of $30^{\circ}$ with the line joining the points (1, 2) and (3, 4), then $\frac{ m _1}{m_2}=$
- A
$2+\sqrt{3}$
- B
$2-\sqrt{3}$
- ✓
$7+4 \sqrt{3}$
- D
$7-4 \sqrt{3}$
AnswerCorrect option: C. $7+4 \sqrt{3}$
(C)
Slope of the line joining the points $(1,2)$ and $(3,4)$ is 1 .
$\tan 30^{\circ}=\left|\frac{ m -1}{1+ m }\right|$
$\Rightarrow \frac{1}{\sqrt{3}}= \pm \frac{m-1}{1+m}$
$\Rightarrow m =\frac{\sqrt{3}+1}{\sqrt{3}-1}$ or $\frac{\sqrt{3}-1}{\sqrt{3}+1}$
$\Rightarrow m =2+\sqrt{3}$ or $2-\sqrt{3}$
$m _1> m _2$
Let $m _1=2+\sqrt{3}, m_2=2-\sqrt{3}$
$\therefore \quad \frac{ m _1}{m_2}=\frac{2+\sqrt{3}}{2-\sqrt{3}}=7+4 \sqrt{3}$
View full question & answer→MCQ 752 Marks
If vertices of a parallelogram are respectively (0, 0), (1, 0), (2, 2) and (1, 2), then angle between diagonals is
- A
$\frac{\pi}{3}$
- ✓
$\frac{\pi}{2}$
- C
$\frac{3 \pi}{2}$
- D
$\frac{\pi}{4}$
AnswerCorrect option: B. $\frac{\pi}{2}$
(B)
Here, Slope of $I ^{ st }$ diagonal $= m _1=\frac{2-0}{2-0}=1$
$\Rightarrow \theta_1=45^{\circ}$
Slope of II ${ }^{ nd }$ diagonal $= m _2=\frac{2-0}{1-1}=\infty$
$\Rightarrow \theta_2=90^{\circ}$
$\Rightarrow \theta_2-\theta_1=45^{\circ}=\frac{\pi}{4}$
View full question & answer→MCQ 762 Marks
If the lines $y=(2+\sqrt{3}) x+4$ and y = kx + 6 are inclined at an angle 60 deg to each other, then the on value of k will be
Answer(C)
$\begin{array}{l}\frac{ k -(2+\sqrt{3})}{1+ k (2+\sqrt{3})}=\sqrt{3} \\ \Rightarrow k -2-\sqrt{3}=\sqrt{3}+2 k \sqrt{3}+3 k \\ \Rightarrow k =\frac{-2(1+\sqrt{3})}{2(1+\sqrt{3})}=-1\end{array}$
View full question & answer→MCQ 772 Marks
The sides AB, BC, CD and DA of a quadrilateral are x + 2y = 3 x = 1 x - 3y = 4 5x + y + 12 = 0 respectively. The angle between diagonals AC and BD is
- A
$45^{\circ}$
- B
$60^{\circ}$
- ✓
$90^{\circ}$
- D
$30^{\circ}$
AnswerCorrect option: C. $90^{\circ}$
(C)
The four vertices on solving are $A (-3,3)$,
$B (1,1), C (1,-1)$ and $D (-2,-2)$.
$m _1=$ slope of $AC =-1$,
$m _2=$ slope of $BD =1$
$m _1 m_2=-1$
Hence, the angle between diagonals AC and BD is $90^{\circ}$.
View full question & answer→MCQ 782 Marks
If $\frac{1}{ ab ^{\prime}}+\frac{1}{ ba ^{\prime}}=0$, then lines $\frac{x}{ a }+\frac{y}{b}=1$ and $\frac{x}{b^{\prime}}+\frac{y}{ a ^{\prime}}=1$ are
AnswerCorrect option: C. Perpendicular to each other
(C)
Slopes of lines $\frac{x}{ a }+\frac{y}{b}=1$
and $\frac{x}{b^{\prime}}+\frac{y}{ a ^{\prime}}=1$ are $\frac{- b }{ a }$ and $\frac{- a ^{\prime}}{ b ^{\prime}}$ respectively
∴ Product of slopes is $\frac{ a ^{\prime} b }{ ab ^{\prime}}$
But $\frac{1}{ ab ^{\prime}}+\frac{1}{ ba ^{\prime}}=0$$\Rightarrow ab ^{\prime}=- a ^{\prime} b$
$\Rightarrow$ Product of slopes $=-1$
Hence option (C) is the correct answer.
View full question & answer→MCQ 792 Marks
If the co-ordinates of the vertices A, B, C of the triangle ABC are (-4, 2), (12, -2) and (8, 6) respectively, then $\angle B$ =
- A
$\tan ^{-1}\left(-\frac{6}{7}\right)$
- B
$\tan ^{-1}\left(\frac{6}{7}\right)$
- C
$\tan ^{-1}\left(-\frac{7}{6}\right)$
- ✓
$\tan ^{-1}\left(\frac{7}{6}\right)$
AnswerCorrect option: D. $\tan ^{-1}\left(\frac{7}{6}\right)$
(D)
Here, equation of AB is
$x+4 y-4=0$ ...(i)
and equation of BC is
$2 x+y-22=0$ ...(ii)
Thus angle between (i) and (ii) is given by$\tan ^{-1}\left(\frac{-\frac{1}{4}+2}{1+\left(-\frac{1}{4}\right)(-2)}\right)=\tan ^{-1} \frac{7}{6}$
View full question & answer→MCQ 802 Marks
Angle between x = 2 and x - 3y = 6 is
MCQ 812 Marks
Angle between the lines $\frac{x}{ a }+\frac{y}{b}=1$ and $\frac{x}{ a }-\frac{y}{b}=1$ is
- ✓
$2 \tan ^{-1} \frac{b}{a}$
- B
$\tan ^{-1} \frac{2 a b}{a^2+b^2}$
- C
$\tan ^{-1} \frac{a^2-b^2}{a^2+b^2}$
- D
AnswerCorrect option: A. $2 \tan ^{-1} \frac{b}{a}$
(A)
The lines are $b x+ a y- ab =0$ and $b x- a y- ab =0$.
Hence, the required angle is
$\theta=\tan ^{-1}\left|\frac{\frac{- b }{ a }-\frac{ b }{ a }}{1-\frac{ b ^2}{ a ^2}}\right|$
$=\tan ^{-1}\left|\frac{2 a b}{b^2-a^2}\right|$
$=2 \tan ^{-1} \frac{b}{ a }$ ...$\ldots\left[\because 2 \tan ^{-1} \frac{y}{x}=\tan ^{-1}\left|\frac{2 x y}{y^2-x^2}\right|\right]$
View full question & answer→MCQ 822 Marks
The angle between the lines $x \cos \alpha+y \sin \alpha= a$ and $x \sin \beta-y \cos \beta= a$ is
AnswerCorrect option: D. $\frac{\pi}{2}-\beta+\alpha$
(D)
Here, $m _1=-\cot \alpha, m _2=\tan \beta$
$\therefore \quad \tan \theta=\left|\frac{-\cot \alpha-\tan \beta}{1-\cot \alpha \tan \beta}\right|$
$\Rightarrow \tan \theta=-\cot (\alpha-\beta)$
$\Rightarrow \theta=\frac{\pi}{2}-\beta+\alpha$
View full question & answer→MCQ 832 Marks
The angle between the lines $x \cos \alpha_1+y \sin \alpha_1= p _1$ and $x \cos \alpha_2+y \sin \alpha_2= p _2$ is
- A
$\alpha_1+\alpha_2$
- ✓
$\alpha_1-\alpha_2$
- C
$2 \alpha_1$
- D
$2 \alpha_2$
AnswerCorrect option: B. $\alpha_1-\alpha_2$
(B)
$\begin{array}{l}\theta=\tan ^{-1}\left|\frac{-\cot \alpha_1+\cot \alpha_2}{1+\cot \alpha_1 \cot \alpha_2}\right| \\ =\tan ^{-1}\left|\frac{\tan \alpha_1-\tan \alpha_2}{1+\tan \alpha_2 \tan \alpha_1}\right|=\alpha_1-\alpha_2\end{array}$
View full question & answer→MCQ 842 Marks
The parallelism condition for two straight lines one of which is specified by the equation ax + by + c = 0, the other being represented parametrically by $x=\alpha t+\beta, y=\gamma t+\delta$ is given by
- A
$\alpha \gamma-b \alpha=0, \beta=\delta=c=0$
- B
$a \alpha- b \gamma=0, \beta=\delta=0$
- ✓
$a \alpha+b \gamma=0$
- D
$a \gamma=b \alpha=0$
AnswerCorrect option: C. $a \alpha+b \gamma=0$
(C)
Given lines are $ar + by + c =0$
and $x=\alpha t +\beta, y=\gamma t +\delta$
After eliminating t , we get
$\gamma x-\alpha y+\alpha \delta-\gamma \beta=0$
For parallelism condition,
$\frac{ a }{\gamma}=\frac{ b }{-\alpha} \Rightarrow a \alpha+ b \gamma=0$
View full question & answer→MCQ 852 Marks
If $P (\alpha, \beta)$ be a point on the line 3x + y = 0 such that the point P and the point Q(1, 1) lie on either side of the line 3x = 4y + 8 then
- A
$\alpha>\frac{8}{15}, \beta<\frac{-8}{5}$
- B
$\alpha<\frac{8}{15}, \beta<\frac{-8}{5}$
- C
$\alpha<\frac{8}{15}, \beta>\frac{-8}{5}$
- D
$\alpha<\frac{8}{15}, \beta>\frac{-8}{5}$
Answer(A)
Points $P (\alpha, \beta)$ and $Q (1,1)$ lie on either side of $L \equiv 3 x-4 y-8=0$.
$\therefore \ L _{(1,1)}=3(1)-4(1)-8=-9<0$
$\Rightarrow 3 \alpha-4 \beta-8>0$
$\Rightarrow 3 \alpha-4(-3 \alpha)-8>0$ ...$\ldots\lfloor y=-3 x\rfloor$
$\Rightarrow \alpha>\frac{8}{15}$
Also, $3 \alpha-4 \beta-8>0$
$\Rightarrow-\beta-4 \beta-8>0 \quad \ldots[3 x=-y]$
$\Rightarrow \beta<-\frac{8}{5}$
View full question & answer→MCQ 862 Marks
A line is such that its segment between the straight lines 5x - y - 4 = 0 and 3x + 4y - 4 = 0 is bisected at the point (1, 5), then its equation is
Answer(A)
Any line through the middle point M(1, 5) of the intercept AB may be taken as
$\frac{x-1}{\cos \theta}=\frac{y-5}{\sin \theta}= r$ ...(i)
Since, the points A and B are equidistant from $M$ and on the opposite sides of it.
Therefore, if the co-ordinates of A are obtained by putting $r = d$ in (i), then the co-ordinates of B are given by putting $r =- d$.
Now, the point $A (1+ d \cos \theta, 5+ d \sin \theta)$ lies on the line $5 x-y-4=0$ and point $B (1- d \cos \theta, 5- d \sin \theta)$ lies on the line $3 x+4 y-4=0$.
$\therefore 5(1+d \cos \theta)-(5+d \sin \theta)-4=0$ and $3(1-d \cos \theta)+4(5-d \sin \theta)-4=0$
Eliminating 'd', we get $\frac{\cos \theta}{35}=\frac{\sin \theta}{83}$
Hence, the required line is $\frac{x-1}{35}=\frac{y-5}{83}$ or
$83 x-35 y+92=0$.
View full question & answer→MCQ 872 Marks
In what direction a line be drawn through the point (1, 2) so that its points of intersection with the line x + y = 4 is at a distance $\frac{\sqrt{6}}{3}$ from the given point
- A
$30^{\circ}$
- B
$45^{\circ}$
- C
$60^{\circ}$
- ✓
$75^{\circ}$
AnswerCorrect option: D. $75^{\circ}$
(D)
Let the required line through the point $(1,2)$ be inclined at an angle 0 to the axis of X . Then its equation is $\frac{x-1}{\cos \theta}=\frac{y-2}{\sin \theta}=r$ ...(i)
The co-ordinates of a point on the line (i) are $(1+r \cos \theta, 2+r \sin \theta)$
If this point is at a distance $\frac{\sqrt{6}}{3}$ from $(1,2)$,then $r =\frac{\sqrt{6}}{3}$
Therefore, the point is $\left(1+\frac{\sqrt{6}}{3} \cos \theta, 2+\frac{\sqrt{6}}{3} \sin \theta\right)$
But this point lies on the line $x+y=4$
$\sin \theta+\cos \theta=\frac{3}{\sqrt{6}}$
$\Rightarrow \frac{1}{\sqrt{2}} \sin \theta+\frac{1}{\sqrt{2}} \cos \theta=\frac{\sqrt{3}}{2}$ ....(Dividing both sides by $\sqrt{2}$ )
$\Rightarrow \sin \left(\theta+45^{\circ}\right)=\sin 60^{\circ}$ or $\sin 120^{\circ}$
$\Rightarrow \theta=15^{\circ}$ or $75^{\circ}$
View full question & answer→MCQ 882 Marks
A line through A(-5, -4) meets the lines x + 3y + 2 = 0 2x + y + 4 = 0 and x - y - 5 = 0 at B, C and D respectively. If $\left(\frac{15}{ AB }\right)^2+\left(\frac{10}{ AC }\right)^2=\left(\frac{6}{ AD }\right)^2$, then the equation of the line is
Answer(A)
The equation of line passing through
$A (-5,-4)$ is $\frac{x+5}{\cos \theta}=\frac{y+4}{\sin \theta}$
Let $AB = r _1, AC = r _2, AD = r _3$
The co-ordinate of B is
$\left(r_1 \cos \theta-5, r_1 \sin \theta-4\right)$
$\therefore \quad r_1=\frac{15}{\cos \theta+3 \sin \theta}$
Similarly, $\frac{10}{ AC }=2 \cos \theta+\sin \theta$ and
$\frac{6}{ AD }=\cos \theta-\sin \theta$
View full question & answer→MCQ 892 Marks
The distance of the line 2x - 3y = 4 from the point (1, 1) measured parallel to the line x + y = 1 is
- A
$\sqrt{2}$
- B
$\frac{5}{\sqrt{2}}$
- C
$\frac{1}{\sqrt{2}}$
- D
View full question & answer→MCQ 902 Marks
The equation of the lines on which the perpendiculars from the origin make $30^{\circ}$ angle with X-axis and which form a triangle of area $\frac{50}{\sqrt{3}}$ with axes, are
- A
$x+\sqrt{3} y \pm 10=0$
- ✓
$\sqrt{3} x+y \pm 10=0$
- C
$x \pm \sqrt{3} y-10=0$
- D
AnswerCorrect option: B. $\sqrt{3} x+y \pm 10=0$
(B)
Let $p$ be the length of the perpendicular from the origin on the given line. Then its equation in normal form is
$x \cos 30^{\circ}+y \sin 30^{\circ}=p \text { or } \sqrt{3} x+y=2 p$
This meets the coordinate axes at $A \left(\frac{2 p }{\sqrt{3}}, 0\right)$ and $B(0,2 p)$
∴ Area of $\triangle OAB =\frac{1}{2}\left(\frac{2 p }{\sqrt{3}}\right) 2 p =\frac{2 p ^2}{\sqrt{3}}$
$\Rightarrow \frac{2 p^2}{\sqrt{3}}=\frac{50}{\sqrt{3}} \Rightarrow p= \pm 5$
Hence, the lines are $\sqrt{3} x+y \pm 10=0$
View full question & answer→MCQ 912 Marks
If we reduce 3x + 3y + 7 = 0 to the form $x \cos \alpha+y \sin \alpha= p$ then the value of p is
- A
$\frac{7}{2 \sqrt{3}}$
- B
$\frac{7}{3}$
- C
$\frac{3 \sqrt{7}}{3}$
- ✓
$\frac{7}{3 \sqrt{2}}$
AnswerCorrect option: D. $\frac{7}{3 \sqrt{2}}$
(D)
Given form is $3 x+3 y+7=0$
$\Rightarrow \frac{3}{\sqrt{3^2+3^2}} x+\frac{3}{\sqrt{3^2+3^2}} y+\frac{7}{\sqrt{3^2+3^2}}=0$
$\Rightarrow \frac{3}{3 \sqrt{2}} x+\frac{3}{3 \sqrt{2}} y=\frac{-7}{3 \sqrt{2}}$
$\therefore \quad p=\left|\frac{-7}{3 \sqrt{2}}\right|=\frac{7}{3 \sqrt{2}}$
View full question & answer→MCQ 922 Marks
An equation of a line whose segment between the coordinate axes is divided by the point $\left(\frac{1}{2}, \frac{1}{3}\right)$ in the ratio 2:3 is
View full question & answer→MCQ 932 Marks
Equation of the line through $(\alpha, \beta)$ which is the midpoint of the line intercepted between the coordinate axes is
- A
$\frac{x}{\alpha}+\frac{y}{\beta}=1$
- B
$\frac{x}{\alpha}+\frac{y}{\beta}=2$
- C
$\frac{x}{\alpha}-\frac{y}{\beta}=-1$
- D
$\frac{x}{\alpha}-\frac{y}{\beta}=-2$
View full question & answer→MCQ 942 Marks
The number of lines that are parallel to 2x + 6y + 7 = 0 and have an intercept of length 10 between the co-ordinate axes is
Answer(B)
The equation of any line parallel to
$2 x+6 y+7=0$ is $2 x+6 y+ k =0$.
This meets the axes at $A \left(-\frac{ k }{2}, 0\right)$ and $B \left(0,-\frac{ k }{6}\right)$
Since, ab = 10
$\Rightarrow \sqrt{\frac{ k ^2}{4}+\frac{ k ^2}{36}}=10 \Rightarrow \sqrt{\frac{10 k ^2}{36}}=10$
$\Rightarrow 10 k ^2=3600 \Rightarrow k = \pm 6 \sqrt{10}$
Hence, there are two lines given by $2 x+6 y \pm 6 \sqrt{10}=0$
View full question & answer→MCQ 952 Marks
If $\left(\frac{3}{2}, \frac{5}{2}\right)$ is the midpoint of line segment intercepted by a line between axes, the equation of the line is
Answer(C)
Let the points of the required line on X -axis and Y -axis be $A ( a , 0)$ and $B (0, b)$ respectively.
Since, $\left(\frac{3}{2}, \frac{5}{2}\right)$ is midpoint of AB.
∴$\frac{ a +0}{2}=\frac{3}{2}$ and $\frac{0+ b }{2}=\frac{5}{2} \Rightarrow a =3$ and $b =5$
∴ The equation of line is $\frac{x}{3}+\frac{y}{5}=1$
$\Rightarrow 5 x+3 y-15=0$
View full question & answer→MCQ 962 Marks
The equation to the line bisecting the join of (3, -4) and (5, 2) and having its intercepts on the X-axis and the Y-axis in the ratio 2: 1 is
Answer(C)
Equation of the line has its intercepts on the X -axis and Y -axis in the ratio $2: 1$
i.e., 2 a and a
∴ $\frac{x}{2 a}+\frac{y}{a}=1
\Rightarrow x+2 y=2 a$ ...(i)
Line (i) also passes through midpoint of $(3,-4)$ and $(5,2)$ i.e.,$(4,-1)$
∴ $4+2(-1)=2 a \Rightarrow a=1$
Hence, the equation of required line is
$x+2 y=2$
View full question & answer→MCQ 972 Marks
For what values of a and b, the intercepts cut off on the coordinate axes by the line ax + by + 8 = 0 are equal in length but opposite in sign to those cut off by the line 2x - 3y + 6 = 0 on the axes?
- A
$a =\frac{8}{3}, b=-4$
- B
$a =-\frac{8}{3}, b=-4$
- C
$a=\frac{8}{3}, b=4$
- ✓
$a=-\frac{8}{3}, b=4$
AnswerCorrect option: D. $a=-\frac{8}{3}, b=4$
(D)
The equation of lines in intercept form are
$\frac{x}{-8 / a}+\frac{y}{-8 / b}=1$
$\frac{x}{-3}+\frac{y}{2}=1$
According to the given condition,
$-\frac{8}{ a }=-(-3)$ and $-\frac{8}{b}=-2$
$\Rightarrow a =-\frac{8}{3}$ and $b =4$
View full question & answer→MCQ 982 Marks
A line L is perpendicular to the line 5x - y = 1 and the area of the triangle formed by the line L and co-ordinate axes is 5. The equation of the line L is
- A
- ✓
$x+5 y= \pm 5 \sqrt{2}$
- C
- D
$x-5 y=5 \sqrt{2}$
AnswerCorrect option: B. $x+5 y= \pm 5 \sqrt{2}$
(B)
A line perpendicular to the line $5 x-y=1$ is given by $x+5 y-\lambda=0= L$
In intercept form $\frac{x}{\lambda}+\frac{y}{\lambda / 3}=1$
So, area of triangle is $\frac{1}{2} \times$ (Multiplication of intercepts)
$\Rightarrow \frac{1}{2}(\lambda) \times\left(\frac{\lambda}{5}\right)=5$
$\Rightarrow \lambda= \pm 5 \sqrt{2}$
Hence, the equation of required line is
$x+5 y= \pm 5 \sqrt{2}$
View full question & answer→MCQ 992 Marks
The equations of the straight line passing through the point (4, 3) and making intercepts on the co-ordinate axes whose sum is 1, is
- ✓
$\frac{x}{2}-\frac{y}{3}=1$ and $\frac{x}{-2}+\frac{y}{1}=1$
- B
$\frac{x}{2}-\frac{y}{3}=-1$ and $\frac{x}{-2}+\frac{y}{1}=-1$
- C
$\frac{x}{2}-\frac{y}{3}=1$ and $\frac{x}{2}+\frac{y}{1}=1$
- D
$\frac{x}{2}+\frac{y}{1}=1$ and $\frac{x}{-2}+\frac{y}{1}=-1$
AnswerCorrect option: A. $\frac{x}{2}-\frac{y}{3}=1$ and $\frac{x}{-2}+\frac{y}{1}=1$
(A)
Here, $a + b =-1$
∴ required line is $\frac{x}{ a }-\frac{y}{1+ a }=1$ ...(i)
Since, line (i) passes through $(4,3)$.
∴$\frac{4}{a}-\frac{3}{1+a}=1$
$\Rightarrow 4+4 a-3 a=a+a^2$
$\Rightarrow a^2=4$
$\Rightarrow a = \pm 2$
$\therefore $ The required lines are $\frac{x}{2}-\frac{y}{3}=1$ and $\frac{x}{-2}+\frac{y}{1}=1$.
View full question & answer→MCQ 1002 Marks
A line passes through the point (3, 4) and cuts off intercepts from the co-ordinates axes such that their sum is 14. The equation of the line is
Answer(B)
Given, $a + b =14 \Rightarrow a =14- b$
Hence, the equation of straight line is $\frac{x}{14- b }+\frac{y}{b}=1$
Also, it passes through $(3,4)$
∴ $\frac{3}{14-b}+\frac{4}{b}=1$
$\Rightarrow b=8 \text { or } 7$
Therefore, equations are $4 x+3 y=24$ and $x+y=7$
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