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11th Maths Weekly Test-3 (Two Dimensional Analytical Geometry)

St. Britto Hr. Sec. School - Madurai

11th Maths Weekly Test-3 (Two Dimensional Analytical Geometry)-Aug 2020

Time : 1.00 Hours

Part - A

Multiple Choice Question - Answer All QuestionChoose...

If p is the length of perpendicular from origin to the line \(\frac{x}{a}+\frac{y}{b}=1\)then

\(\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}\)
\(\frac{1}{p^2}=\frac{1}{a^2}-\frac{1}{b^2}\)
\(\frac{1}{p^2}=-\frac{1}{a^2}+\frac{1}{b^2}\)
\(\frac{1}{p^2}=-\frac{1}{a^2}-\frac{1}{b^2}\)

The intercepts of the perpendicular bisector of the line segment joining (1, 2) and (3,4) with coordinate axes are

5,-5
5,5
5,3
5,-4

The equation of the straight line which passes through the point (2,4) and have intercept on the axes equal in magnitude but opposite in sign is

x-y=2
x-y+2=0
x-y+1=0
x-y-1=0

The points (a,0),(0,b) and (1,1) will be collinear if

a+b=1
a+b=2
\(\frac{1}{a}+\frac{1}{b}=1\)
a+b=0

The image of the point (2, 3) in the line y = -x is

(-3, -2)
(-3,2)
(-2, -3)
(3,2)

If the co-ordinates of a variable point p be \((t+\frac{1}{t},t-\frac{1}{t})\)where t is the parameter then the locus of p

xy=1
x²+y²=4
x²-y²=4
x²-y²=8

Find the nearest point on the line 3x + y = 10 from the origin is:

(2, 1)
(1, 2)
(3, 1)
(1,3)

The equation of the line passing through (1, 5) and perpendicular to the line 3x -5y + 7 = 0 is

5x+3y-20=0
3x-5y+7=0
3x-5y+6=0
5x+3y+7=0

The equation of the locus of the point whose distance from y-axis is half the distance from origin is

x²+3y=0
x²-3y²=0
3x²+y²=0
3x²-y²=0

The image of the point (1,2) with respect to the line y=x is

(-1,-2)
(2,1)
(2,-1)
(2,1)

The point on the line 2x- 3y = 5 is equidistance from (1,2) and (3,4) is

(7,3)
(4,1)
(1,-1)
(-2,3)

Which one of the following statements is false?

The image of a point \((\alpha\beta)\) about x-axis \((\alpha,-\beta)\)
The image of the line ax+by+c=0 about x-axis is ax-by+c=0
The image of a point \((\alpha,\beta)\) about y-axis \((-\alpha,\beta)\)
The image of the line ax+by+c=0 about y-axis is ax-by+c=0

The slope of the line joining A and B where A is (-1, 2) and B is the point of intersection of the lines 2x + 3y = 5 and 3x + 4y = 7 is:

-2
2
\(\frac{1}{2}\)
-\(\frac{1}{2}\)

The condition that the slope of one of the lines represented by ax²+2hxy+by²=0 is n times the slope of the other is

4nh²=ab(1+n)²
8h²=9ab
4n=ab(1+n)²
4nh²=ab

If a vertex of a square is at the origin and its one side lies along the line 4x + 3y - 20 = 0, then the area of the square is

20 sq. units
16 sq. units
25 sq. units
4 sq.units

The line (p + 2q)x + (p - 3q)y = p - q for different values of p and q passes through the point

\(\left(\frac{3}{5},\frac{5}{2}\right)\)
\(\left(\frac{2}{5},\frac{2}{5}\right)\)
\(\left(\frac{3}{5},\frac{3}{5}\right)\)
\(\left(\frac{2}{5},\frac{3}{5}\right)\)

The length of perpendicular from the origin to a line is 12 and the line makes an angle of 120° with the positive direction of y-axis. then the equation of line is

\(x+y\sqrt 3=24\)
\(x+y=12\sqrt 2\)
x+y=24
\(x+y=12\sqrt 3\)

The equation x²+kxy+y²-5x-7y+6=0 represents a pair of straight lines then k=

\(\frac{5}{3}\)
\(\frac{10}{3}\)
\(\frac{3}{2}\)
\(\frac{3}{10}\)

If the equation of the base opposite to the vertex (2,3) of an equilateral triangle is x +y = 2, then the length of a side is

\(\sqrt{\frac{3}{2}}\)
6
\(\sqrt{6}\)
\(3\sqrt{2}\)

The distance between the line 12x - 5y + 9 = 0 and the point (2, 1) is

\(\pm\frac{28}{13}\)
\(\frac{28}{13}\)
\(-\frac{28}{13}\)
none of these

If the two straight lines x + (2k -7)y + 3 = 0 and 3kx + 9y - 5 = 0 are perpendicular then the value of k is

k=3
\(k=\frac13\)
\(k=\frac23\)
\(k=\frac32\)

AB = 12 cm. AB slides with A on x-axis, B on y-axis respectively. Then the radius of the circle which is the locus of ΔAOB, where O is origin is:

36
4
16
9

Separate equation of lines for a pair of lines whose equation is x²+xy-12y²=0 are

x+4y=0 and x+3y=0
2x-3y=0 and x-4y=0
x-6y=0 and x-3y=0
x+4y=0 and x-3y=0

If one of the lines of my²+(1-m²)xy-mx²=0 is a bisector of the angle between the lines xy=0 then m is

\(\frac{-1}{2}\)
-2
1
2

\(\theta\) is acute angle between the lines x²-xy- 6y² = 0, then \(\frac{2\cos\theta+3\sin\theta}{4\sin\theta+5\cos\theta}\) is

1
\(-\frac{1}{9}\)
\(\frac{5}{9}\)
\(\frac{1}{9}\)

Pair of lines perpendicular to the lines represented by ax²+2hxy+by²=0 and through origin is

ax²+2hxy+by²=0
bx²+2hxy+ay²=0
bx²-2hxy+ay²=0
bx²-2hxy+ay²=0

If O is the origin and Q is a variable point on y²=x, then the locus of the mid-point of OQ is

y²=2x
2y²=x
4y²=x
y=2x²

The locus of a point which is collinear with the points (a,0) and (0,b) is

x+y=1
\(\frac{x}{a}+\frac{y}{b}=1\)
x+y=ab
\(\frac{x}{a}-\frac{y}{b}=1\)

The equation 3x²+2hxy+3y²=0 represents a pair of straight lines passing through the origin. The two lines are

real and distinct if h²>3
real and distinct if h²>0
real and distinct h²>6
real and distinct if h²-9=0

A line perpendicular to the line 5x - y = 0 forms a triangle with the coordinate axes. If the area of the triangle is 5 sq. units, then its equation is

\(x+5y\pm5\sqrt2=0\)
\(x-5y\pm5\sqrt2=0\)
\(5x+y\pm5\sqrt2=0\)
\(5x-y\pm5\sqrt2=0\)

The locus of a point which moves such that it maintains equal distances from two fixed points is a

straight line
line bisector
pair of straight lines
angle bisector

The locus of a moving point P(a cos³θ, a sin³θ) is

\({ x }^{ \frac { 2 }{ 3 } }+{ y }^{ \frac { 2 }{ 3 } }={ a }^{ \frac { 2 }{ 3 } }\)
x²+y²=a²
x + y = a
\({ x }^{ \frac { 3 }{ 2 } }+{ y }^{ \frac { 3 }{ 2 } }={ a }^{ \frac { 3 }{ 2 } }\)

The equating straight line with y-intercept -2 and inclination with x-axis is 135° is:

x+y-2=0
y-x+2=0
y+x+2=0
none

The equation of the straight line bisecting the line segment joining the points (2,4) and (4,2) and making an angle of 45⁰ with positive direction of x-axis is

x+y=6
x-y=0
x-y=6
x+y=0

The angle between the lines 2x - y + 3 = 0 and x + 2y + 3 = 0 is

90°
60°
45°
30°

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