Mathematics Class XI

Question 1 of 25

1. Question
1 point(s)
Find the equation of the line passing though the point (5,2) and prependicular to the line joining the point (2,3) & (3,-1) is
- x – 4y – 3 = 0
- x – 4y + 3 = 0
- x + 4y + 3 =0
- x + 4y – 3 =0
Correct

Incorrect

Hint

Slope of the line passing through
(2,3) & (3,-1) is $\dfrac{{ - 1 - 3}}{{3 - 2}} = - 4$
$\therefore {m_1} = - 4$
slope of required line $\left( {{m_2}} \right) = \dfrac{1}{4}$
since, the equation of the line passing through (5,2) & having slope $\dfrac{1}{4}$ is $y - 2 = \dfrac{1}{4}\left( {x - 5} \right)$
$\Rightarrow 4y - 8 = x - 5$
$\Rightarrow x - 4y + 3 = 0$
Question 2 of 25

2. Question
1 point(s)
Find the points on the line $x + y = 4$, which lie t a unit distance from the line $4x + 3y = 10$
- (3,1) & (7,11)
- (3,1) & (-7,11)
- (-3,1) & (-7,11)
- (-3,1) & (7,11)
Correct

Incorrect

Hint

Given $x + y =4$ , Let P(h,k) lie on this line $\Rightarrow h + k = 4$
The distance of the point P(h,k) from the line 4x + 3y = 10 is
$d = \left| {\dfrac{{4h + 3k - 10}}{{\sqrt {16 + 9} }}} \right| \Rightarrow 1 = \left| {\dfrac{{4h + 3k - 10}}{5}} \right|$
$\Rightarrow 4h + 3k - 10 = \pm 5$
Taking positive sign,
$4h + 3k = 15 \Rightarrow k = 1\& h = 3$
$\therefore P\left( {h,k} \right) = \left( {3,1} \right)$
Taking negative sign,
$4h + 3k = 5.\left( {h = 4 - k} \right)$
$\Rightarrow k = 11,h = - 7\therefore P\left( {h,k} \right) = \left( { - 7,11} \right)$
$\therfore$ The point are (3,1) & (-7,11).
Question 3 of 25

3. Question
1 point(s)
The tangent of an angle between the lines $\dfrac{x}{a} + \dfrac{y}{b} = 1\& \dfrac{x}{a} – \dfrac{y}{b} = 1$ is
- $\dfrac{{2ab}}{{{a^2} + {b^2}}}$
- $\dfrac{{2ab}}{{{a^2} – {b^2}}}$
- $\dfrac{{2ab}}{{{b^2} – {a^2}}}$
- $\dfrac{{ab}}{{{a^2} + {b^2}}}$
Correct

Incorrect

Hint

$\dfrac{x}{a} + \dfrac{y}{b} = \& \dfrac{x}{a} - \dfrac{y}{b} = 1,slope\left( {{m_2}} \right) = \dfrac{b}{a}$
$\therefore \tan \theta = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|$
$\Rightarrow \tan \theta \left| {\dfrac{{ - \dfrac{b}{a} - \dfrac{b}{a}}}{{1 + \left( {\dfrac{{ - b}}{a}} \right)\left( {\dfrac{b}{a}} \right)}}} \right|$
$= \left| {\dfrac{{\dfrac{{ - 2b}}{a}}}{{1 - \dfrac{{{b^2}}}{{{a^2}}}}}} \right| = \left| {\dfrac{{2ab}}{{{a^2} - {b^2}}}} \right|$
$\Rightarrow \tan \theta = \dfrac{{2ab}}{{{a^2} - {b^2}}}$
Question 4 of 25

4. Question
1 point(s)
Find the angle between the lines $y = \left( {2 – \sqrt 3 } \right)\left( {x + 5} \right)\& y = \left( {2 + \sqrt 3 } \right)\left( {x – 7} \right)$
- $\pi /4$
- $\pi/3$
- $\pi/6$
- $\pi/8$
Correct

Incorrect

Hint

$y = \left( {2 - \sqrt 3 } \right)\left( {x + 5} \right)$
$slope\left( {{m_1}} \right) = 2 - \sqrt 3$
$\& y = \left( {2 + \sqrt 3 } \right)\left( {x - 7} \right)$
$slope\left( {{m_2}} \right) = \left( {2 + \sqrt 3 } \right)$
$\therefore \tan \theta = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|$
$= \left| {\dfrac{{\left( {2 - \sqrt 3 } \right) - \left( {2 + \sqrt 3 } \right)}}{{1 + \left( {2 - \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)}}} \right| = \left| {\dfrac{{ - 2\sqrt 3 }}{{1 + 4 - 3}}} \right|$
$\Rightarrow \tan \theta = \left| { - \sqrt 3 } \right| = \sqrt 3$
$\Rightarrow \theta = {\tan ^{ - 1}}\sqrt 3 = \pi /3$
Question 5 of 25

5. Question
1 point(s)
Find the equation of the line passing through the points (7,-9) and (5,-7) is
- $x + y + 3 = 0$
- $x + y – 1= 0$
- $x + y + 2 =0$
- $x + y – 2 =0$
Correct

Incorrect

Hint

$y - {y_1} = \left( {\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right)\left( {x - {x_1}} \right)\left( {two - po{{\text int}} form} \right)$
$\Rightarrow y + 9 = \left( {\dfrac{{ - 7 + 9}}{{5 - 7}}} \right)\left( {x - 7} \right)$
$y - {y_1} = \left( {\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right)\left( {x - {x_1}} \right)\left( {two - po{{\text int}} form} \right)$
$\Rightarrow y + 9 = \left( {\dfrac{2}{{ - 2}}} \right)\left( {x - 7} \right)$
$\Rightarrow y + 9 = - \left( {x - 7} \right)$
$\Rightarrow x + y + 9 - 7 = 0$
$\Rightarrow x + y + 2 = 0$
Question 6 of 25

6. Question
1 point(s)
Find the equation of the line passing through (1,2) and making angle $45^\circ $ with y-axis in positive direction.
- $x + y – 1 =0$
- $x – y – 1 = 0$
- $x – y + 1 = 0$
- $x + y + 1 = 0$
Correct

Incorrect

Hint

Given, angle with y-axis= $45^\circ$
$\Rightarrow$ angle with x-axis= $45^\circ$
slope (m)= tan $45^\circ =1$
so, equation of line passing through (1,2) & having slope (m) 1, is y-2=1(x-1)
$\Rightarrow y - 2 = x - 1 \Rightarrow x - y + 2 - 1 = 0$
$\Rightarrow x - y + 1 = 0$
Question 7 of 25

7. Question
1 point(s)
Slope of a line which cuts off intercepts of equal lengths on the axes is
- -1
- 0
- 2
- 1
Correct

Incorrect

Hint

Equation of line is
$\dfrac{x}{a} + \dfrac{y}{b} = 1$ (Intercept from)
$\Rightarrow \dfrac{x}{a} + \dfrac{y}{b} = 1\left( {a = b} \right)$
$\Rightarrow x + y = a$
$\Rightarrow y = - x + a$
$\therefore$ slope (m) = – 1
Question 8 of 25

8. Question
1 point(s)
If P is the length of perpendicular from the origin on the line $\dfrac{x}{a} + \dfrac{y}{b} = 1$ and ${a^2},{p^2}\& {b^2}$ are in A.P, then
- ${a^4} – {b^4} = 0$
- ${a^4} + {a^4} =0$
- ${a^4} – {2b^4} =0$
- ${a^4} + {2b^4} =0$
Correct

Incorrect

Hint

Equation of the line is $\dfrac{x}{a} + \dfrac{y}{b} = 1$
perpendicular length from origin on the above line is given by p.
i.e. P= $\dfrac{1}{{\sqrt {\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}}} }} = \dfrac{{ab}}{{\sqrt {{a^2} + {b^2}} }}$
$\Rightarrow {p^2} = \dfrac{{{a^2}{b^2}}}{{{a^2} + {b^2}}}$
Given that ${a^2},{p^2}\& {b^2}$ are in A.P $2{P^2} = {a^2} + {b^2}$
$\Rightarrow \dfrac{{2{a^2}{b^2}}}{{{a^2} + {b^2}}} = {a^2} + {b^2}$
$\Rightarrow 2{a^2}{b^2} = {({a^2} + {b^2})^2} = {a^4} + {b^4} + 2{a^2}{b^2}$
$\Rightarrow {a^4} + {b^4} = 0$
Question 9 of 25

9. Question
1 point(s)
A line cutting off intercept -3 from the y-axis and the tangent at angle to the x-axis is $\dfrac{3}{5}$, its equation is
- $5x – 3y + 15 =0$
- $3x – 5y – 15 =0$
- $3x – 5y + 15 = 0$
- $5x – 3y – 15 =0$
Correct

Incorrect

Hint

Given that, c=-3 & m = $\dfrac{3}{5}$
$\therefore$ Equation of the line is $y = mx + c$
$\Rightarrow y = \dfrac{3}{5}x - 3$
$\Rightarrow 5y = 3x - 15$
$\Rightarrow 3x - 5y - 15 = 0$
Question 10 of 25

10. Question
1 point(s)
The equation of the straight line passing through the point (3,2) & perpendicular to the line y=x is
- $x – y = 5$
- $x + y = 5$
- $x + y = 1$
- $x – y = 1$
Correct

Incorrect

Hint

Line passes through (3,2) & perpandicular to the line y=x
$\therefore$ slope (m) = -1 [ line is perpendicular to line y=x]
$\therefore$ Equation of the line is
y – 2 = -1 (x – 3)
$\Rightarrow y - 2 + x - 3 = 0$
$\Rightarrow x + y - 5 = 0$
$\Rightarrow x + y = 5$
Question 11 of 25

11. Question
1 point(s)
For what values of a & b the intercepts cut off on the cordinate axes by the line $ax + by + 8 = 0$ are equal length but opposite in signs to those cut off by the line $2x – 3y + 6 = 0$ on the axes ?
- $\left( {\dfrac{{ – 8}}{3}, – 4} \right)$
- $\left( {\dfrac{{ – 8}}{3},4} \right)$
- $\left( {\dfrac{8}{3},4} \right)$
- none of these
Correct

Incorrect

Hint

Given, $ax + by + 8 = 0$
$\Rightarrow \dfrac{x}{{ - 8/a}} + \dfrac{y}{{ - 8/b}} = 1$ so, the intercepts are $\dfrac{{ - 8}}{a},\dfrac{{ - 8}}{b}$
Another given equation of line is $2x - 3y + 6 = 0$
$\Rightarrow \dfrac{x}{{ - 3}} + \dfrac{y}{2} = 1$ , so the intercepts are -3, 2 According to the question.
$\dfrac{{ - 8}}{a} = 3\& \dfrac{{ - 8}}{b} = - 2$
$\Rightarrow a = \dfrac{{ - 8}}{3}\& b = 4$
Question 12 of 25

12. Question
1 point(s)
If the intercept of a line between the coordinate axes is divided by the point (-5,4) in the ratio 1:2, then find the equation of the line?
- $8x – 5y + 60 = 0$
- $8x – 5y – 60 =0$
- $8x + 5y + 60 = 0$
- $8x + 5y – 60 = 0$
Correct

Incorrect

Hint

Let intercepts are h, k
The coordnates of A & B are (h,k) & (0,k) respectively
$\therefore - 5 = \dfrac{{1 \times 0 + 2 \times h}}{{1 + 2}} \Rightarrow - 5 = \dfrac{{2h}}{3} \Rightarrow h = \dfrac{{15}}{2}$
$\& 4 = \dfrac{{1.k + 0.2}}{{1 + 2}} \Rightarrow k = 12$
$\therefore A = \left( {\dfrac{{ - 15}}{2},0} \right)\& B = \left( {0,12} \right)$
$\therefore$ Equation of the line AB is
$y - 0 = \dfrac{{12 - 0}}{{0 + \dfrac{1}{2}}}\left( {x + \dfrac{{15}}{2}} \right)(two - po{{\text int}} form)$
$\Rightarrow y = \dfrac{{12 \times 2}}{{15}}\left( {x + \dfrac{{15}}{2}} \right)$
$\Rightarrow 5y = 8x + 60 \Rightarrow 8x - 5y + 60 = 0$
Question 13 of 25

13. Question
1 point(s)
Find the equation of a line on which length of perpendicular from the orign is four units and the line makes an angle of $120^\circ $ with the positive direction of x-axis.
- $\sqrt {3}x – y – 8 = 0$
- $\sqrt {3}x + y + 8 = 0$
- $\sqrt {3}x + y – 8 = 0$
- $\sqrt {3}x – y + 8 = 0$
Correct

Incorrect

Hint

Given, OC= P= 4 Units
$m \angle BAX=120^\circ$
Let $m \angle COA = \alpha ,m \angle OCA = 90^\circ$
$\Rightarrow m \angle BAX = m \angle COA + m \agle OCA$
$\Rightarrow 120^\circ = \alpha + 90^\circ$
$\Rightarrow \alpha = 30^\circ$
Thus, equation of line is
$x\cos 30^\circ + y\sin 30^\circ = 4$
$\Rightarrow x \cdot \frac{{\sqrt 3 }}{2} + y \cdot \frac{1}{2} = 4$
$\Rightarrow \sqrt 3 x + y = 8$
$\Rightarrow \sqrt 3 x + y - 8 = 0$
Question 14 of 25

14. Question
1 point(s)
If the equation of the base of an equilateral triangle is x+y=2 & the vector is (2,-1) then find the length of the side of the triangle.
- $\sqrt{\dfrac 32}$
- $\dfrac{\sqrt{3}}{2}$
- $\dfrac2{\sqrt{3}}$
- $\sqrt{\dfrac 23}$
Correct

Incorrect

Hint

Equation of the base is x+y=2.
In $\triangle ABD$ , $\sin 60^\circ = \dfrac{{AD}}{{AB}}$
$\Rightarrow AD = AB\sin 60^\circ = AB\dfrac{{\sqrt 3 }}{2}$
$\therefore AD = AB\dfrac{{\sqrt 3 }}{2}\left( 1 \right)$
Length of perpendicular from (2,-1) to the
line x+y=2 is given by
$AD = \left| {\dfrac{{2 + \left( { - 1} \right) - 2}}{{\sqrt {{1^2} + {1^2}} }}} \right| = \dfrac{1}{{\sqrt 2 }}$
From (i), $\dfrac{1}{{\sqrt 2 }} = AB\dfrac{{\sqrt 3 }}{2}$
$\Rightarrow AB = \sqrt {\dfrac{2}{3}}$
Question 15 of 25

15. Question
1 point(s)
A variable line passes through a fixed point P. The algebraic sum of the perpendiculr drawn from the points (2,0), (0,2) & (1,1) on the line is zero. Find the coordinates of the point P.
- (1,2)
- (-1,1)
- (1,1)
- (2,1)
Correct

Incorrect

Hint

Slope=m, P=(x,y)
Eqution of line is $y - {y_1} = m\left( {x - {x_1}} \right)$ (Slope-point form)
Given A(2,0), B(0,2) & C(1,1)
Perpendicular distance from A is
$\dfrac{{0 - {y_1} - m\left( {2 - {x_1}} \right)}}{{\sqrt {1 + {m^2}} }}$
perpendicular distance from B is $\dfrac{{2 - {y_1} - m\left( {0 - {x_1}} \right)}}{{\sqrt {1 + {m^2}} }}$
Perpendicular distance from C is $\dfrac{{1 - {y_1} - m\left( {1 - {x_1}} \right)}}{{\sqrt {1 + {m^2}} }}$
Now, $\dfrac{{ - {y_1} - 2m + m{x_1} + 2 - {y_1} + m{x_1} + 1 - {y_1} - m + m{x_1}}}{{\sqrt {1 + {m^2}} }} = 0$
$\Rightarrow - 3{y_1} - 3m + 3m{x_1} + 3 = 0 \Rightarrow - {y_1} - m + m{x_1} + 1 = 0$
Since, (1,1) lies on this line. P=(1,1)
Question 16 of 25

16. Question
1 point(s)
A straight line moves so that sum of the reciprocals of its intercepts mde on axes is constant. So the line passes through a fixed point.
- (k,-k)
- (-k,k)
- (k,k)
- (-k,2k)
Correct

Incorrect

Hint

$\dfrac{x}{a} + \dfrac{y}{b} = 1$ ( intercept form)
Given $\dfrac{1}{a} + \dfrac{1}{b} =$ Constant
$\Rightarrow \dfrac{1}{a} + \dfrac{1}{b} = \dfrac{1}{k}$ (say)
$\Rightarrow \dfrac{k}{a} + \dfrac{k}{b} = 1$
So (k,k) lies on $\dfrac{x}{a} + \dfrac{y}{b} = 1$
Hence, the line passes through the
fixed point (k,k).
Question 17 of 25

17. Question
1 point(s)
If the sum of distance of a moving point in a plane from the axes is 1, then find the locus of the point.
- 2x+y=1
- x-y=1
- 2y-x=1
- x+2y=1
Correct

Incorrect

Hint

Let P be moving point of coordinates (x,y).
Given, sum of the distances of this point
in a plane from the xes is 1.
$\left| x \right| + \left| y \right| = 1$
$\Rightarrow \pm x \pm y = 1$
$\Rightarrow x + y = 1,x - y = 1$
$- x - y = 1, - x + y = 1$
These are the equations of given locus. x-y=1
Question 18 of 25

18. Question
1 point(s)
The tangent of angle between the lines whose intercepts on the axes are a,-b & b, -a respectively, is
- $\dfrac{{{a^2} – {b^2}}}{{ab}}$
- $\dfrac{{{b^2} – {a^2}}}{{ab}}$
- $\dfrac{{{b^2} – {a^2}}}{{2ab}}$
- none of these
Correct

Incorrect

Hint

x-intercept=a, y-intercept=-b
Equation of the line is $\dfrac{x}{a} - \dfrac{y}{b} = 1$
$\Rightarrow \dfrac{y}{b} = \dfrac{x}{a} - 1 \Rightarrow y = \dfrac{{bx}}{a} - b$
$\therefore {m_1} = \dfrac{b}{a}$
x-intercept=b, y-intercept=-a
Eqution of the line is $\dfrac{x}{b} - \dfrac{y}{a} = 1$
$\Rightarrow \dfrac{y}{a} = \dfrac{x}{b} - 1 \Rightarrow y = \dfrac{a}{b}x - a$
$\therefore {m_2} = \dfrac{a}{b}$
$\tan \theta = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right| = \left| {\dfrac{{b/a - a/b}}{{1 + \dfrac{a}{b} \cdot \dfrac{b}{a}}}} \right| = \left| {\dfrac{{\dfrac{{{b^2} - {a^2}}}{{ab}}}}{2}} \right|$
$= \left| {\dfrac{{{b^2} - {a^2}}}{{2ab}}} \right| = \dfrac{{{b^2} - {a^2}}}{{2ab}}$
Question 19 of 25

19. Question
1 point(s)
The distance between two lines $y = mx + {c_1}$& $y = mx + {c_2}$ is
- $\dfrac{{{c_1} – {c_2}}}{{\sqrt {{m^2} + 1} }}$
- $\dfrac{{\left| {{c_1} – {c_2}} \right|}}{{\sqrt {{m^2} + 1} }}$
- $\dfrac{{{c_2} – {c_1}}}{{\sqrt {{m^2} + 1} }}$
- 0
Correct

Incorrect

Hint

Given, $y = mx + {c_1}$
$y = mx + {c_2}$
These two lines are parallel lines
$\therefore d = \dfrac{{\left| {{c_1} - {c_2}} \right|}}{{\sqrt {{m^2} + 1} }}$
Question 20 of 25

20. Question
1 point(s)
If the co-ordinates of a middle point of the portion of a line intercepted between the co-ordinate axes is (3,2), then the equation of the line will be
- 2x+3y=12
- 3x+2y=12
- 4x-3y=6
- 5x-2y=10
Correct

Incorrect

Hint

P=(3,2)
$3 = \dfrac{{1 \cdot 0 + 1 \cdot a}}{{1 + 1}}$
$\Rightarrow 3 = \dfrac{a}{2} \Rightarrow a = 6$
$2 = \dfrac{{1 \cdot b + 1 \cdot 0}}{{1 + 1}} \Rightarrow 2 = \dfrac{b}{2} \Rightarrow b = 4$
Thus, equation of line is
$\dfrac{x}{6} + \dfrac{y}{4} = 1$ (intercept form)
$\Rightarrow \dfrac{{2x + 3y}}{{12}} = 1$
$\Rightarrow 2x + 3y = 12$
Question 21 of 25

21. Question
1 point(s)
Equation of diagonals of the square formed by the lines x=0, y=0, x=1 & y=1 are
- y=x, y+x=1
- 2y=x, y+x=1/3
- y=x, y+2x=1
- y=2x, y+2x=1
Correct

Incorrect

Hint

Equation OB is $y - 0 = \left( {\dfrac{{1 - 0}}{{1 - 0}}} \right)\left( {x - 0} \right)(Two-point form)$
$\Rightarrow y = x\left( 1 \right)$
Equation of AC is $y - 0 = \left( {\dfrac{{1 - 0}}{{0 - 1}}} \right)\left( {x - 1} \right)$
$\Rightarrow y = - x + 1$
$\Rightarrow y + x = 1\left( 2 \right)$
Question 22 of 25

22. Question
1 point(s)
For specifying a straight line, how many geometrical parameter should be known?
- 1
- 2
- 3
- 4
Correct

Incorrect

Hint

Equation of straight line are
$\left( i \right)y = mx + c$ , parameter=2
$\left( {ii} \right)\dfrac{x}{a} + \dfrac{y}{b} = 1$ , parameter=2
$\left( {iii} \right)y - {y_1} = m\left( {x - {x_1}} \right)$ ,parameter=2
$\left( {iv} \right)x\cos \alpha + y\sin \alpha = P$ , parameter=2
Question 23 of 25

23. Question
1 point(s)
The equation of the line through the intersection of the lines 2x-3y=0 & 4x-5y=2 and through the point (2,1) is
- x+y+1=0
- x+y-1=0
- x-y-1=0
- x-y+1=0
Correct

Incorrect

Hint

Given, 2x-3y=0 & 4x-5y=2
Solving above equation we get x=3, y=2
(x,y)=(3,2) is intersection point
Equation of the line passes through(3,2) & (2,1) is
$y - 2 = \left( {\dfrac{{1 - 2}}{{2 - 3}}} \right)\left( {x - 3} \right)$ (Two-point form)
$\Rightarrow y - 2 = \left( {x - 3} \right)$
$\Rightarrow x - y - 1 = 0$
Question 24 of 25

24. Question
1 point(s)
The equation of the line through point (3,2) & perpendicular to the line x+2y+1=0
- 2x-y=4
- 2x+y=4
- x-2y=4
- x+2y=4
Correct

Incorrect

Hint

Point=(3,2), line is x+2y+1=0
Slope(m)=2
Thus, equation of the required line is y-2=2(x-3)
$\Rightarrow y - 2 = 2x - 6$
$\Rightarrow 2x - y = 4$
Question 25 of 25

25. Question
1 point(s)
The equation of the line through the intersection of the lines 2x-3y=0 & 4x-5y=2 and inclined equally to the axes is
- x+y=3
- x+y=-3
- x+y=5
- x+y=-5
Correct

Incorrect

Hint

Given, 2x-3y=0 & 4x-5y=2
Solving above two equations we get x=3, y=2
Since, the line equally inclined to
coordinated axes, then
$m = \pm \tan 45^\circ = \pm 1$
Taking negative value, then equation of line is
y-2=-1(x-3)
x+y=5

Mathematics Class XI

Practice Questions 3-Straight lines-Class XI

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