Straight Lines Test -4

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Straight Lines Test -4

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Question 1
1 / -0

The vertices of a triangle are ( 0 , 3 ) , ( - 3 , 0 ) and ( 3 , 0 ) . The orthocenter of the triangle is

A
( 3 , 0 )

B
( 0 , 3 )

C
( - 3 , 0 )

D
none of these.

Solution

From the given points it is clear that these points form a right angled isosceles triangle.

The orthocentre of all right angle lies on its right angle vertex.

Here the right angle vertex is (0,3)

Hence the orthocentre is (0,3)
Question 2
1 / -0

The lines y = mx , y + 2x = 0 , y = 2x + λ and y = - mx + λ form a rhombus if m =

A
- 1

B
1

C
-2

D
none of these

Solution

Rhombus is a parallelogram in which the opposite sides are equaland parallel.

Therefore the lines y = mx and y = -2x are parallel, similarly y = 2x+ λ and y = -mx + λ are parallel.

If two lines are equal, then their slopes are qual

This implies m = -2
Question 3
1 / -0

Two points ( a , 0 ) and ( 0 , b ) are joined by a straight line. Another point on this line is

A
( - 3a , 2b )

B
( a , - 2b )

C
( 3a , - 2b)

D
none of these

Solution

The slope of the line joining the points (a,0) and (0,b) is [b-0]/[0-a] = -(b/a)

Hence the equation of the line is y = (-b/a)x +b

i.e; ay = -bx +ab

Substituting the x coordinate 3a in the place of x in the above equation we get y = -2b

Hence (3a,-2b) is another point on the line.
Question 4
1 / -0

The coordinates of the foot of perpendicular from ( 0 , 0 ) upon the line x + y = 2 are

A
( 1, 2 )

B
( 1, 1 )

C
( - 1 , 2 )

D
none of these

Solution

The equation of the line perpendicular to the given line is x - y + k = 0

Since it passes through the origin,

0 - 0 + k = 0

Therefore k = 0

Hence the equation of the line is x - y = 0

On solving these two equations we get x = 1 and y = 1

The point of intersection of these two lines is (1,1)

Hence the coordinates of the foot of the perpendicular is (1,1)
Question 5
1 / -0

The area of the triangle whose sides are along the lines x = 0 , y = 0 and 4x + 5y = 20 is

A
10

B
20

C
1/10

D
none of these

Solution

The equation 4x + 5y = 20 can be written as x/5+y/4 = 1

This implies the intercepts cut by this line on the X and Y axes are 5 and 4 respectively.

Hence the area of the triangle is 1/2 [ 5 x 4] = 10 square units
Question 6
1 / -0

Point $$P(8,5)$$ is the midpoint of segment $$AB$$, Co-ordinates of $$A$$ are $$(5,y)$$ and $$B$$ are $$(x,-3)$$. What is the value of $$x$$?

A
$$-11$$

B
$$11$$

C
$$7$$

D
$$-7$$

Solution

$$\cfrac { 5+x }{ 2 } =8\Rightarrow x=11\quad $$
Question 7
1 / -0

Statement-I : Inclination and Slope of a Line give an indication of the Length of that Line.
Statement-II : Inclination and Slope of a Line give an indication of the Direction of that Line.
Choose the correct Alternative.

A
Only Statement I is True.

B
Only Statement II is True.

C
Both Statement I and Statement II are True.

D
Both Statement I and Statement II are False.

Solution

Inclination of a line is the angle it makes with positive x axis and hence gives direction of the line.
Question 8
1 / -0

The distance of point A$$(-2, 3, 1)$$ from the line PQ through P$$(-3, 5, 2)$$, which makes equal angles with the axes is?

A
$$2/\sqrt{3}$$

B
$$\sqrt{14/3}$$

C
$$16/\sqrt{3}$$

D
$$5/\sqrt{3}$$

Solution

Given,
The line $$PQ$$ makes equal angle with axes, say $$\alpha$$.
So the direction cosine of the line will be $$(\cos { \alpha } ,\cos { \alpha } ,\cos { \alpha } )$$
Also,
$$\cos { ^{ 2 }\alpha } +\cos { ^{ 2 }\alpha } +\cos { ^{ 2 }\alpha } =1\\ \cos { \alpha } =\cfrac { 1 }{ \sqrt { 3 } } $$
$$\therefore$$ the direction cosine of the line is $$(\cfrac { 1 }{ \sqrt { 3 } } ,\cfrac { 1 }{ \sqrt { 3 } } ,\cfrac { 1 }{ \sqrt { 3 } } )$$
Next,we move on to find the projection of $$AP$$ on $$PQ$$
$$PD=(-2+3)\cfrac { 1 }{ \sqrt { 3 } } +(3-5)\cfrac { 1 }{ \sqrt { 3 } } +(1-2)\cfrac { 1 }{ \sqrt { 3 } } =\cfrac { -2 }{ \sqrt { 3 } } \\ \therefore { AD }^{ 2 }={ AP }^{ 2 }-{ PD }^{ 2 }={ (3-2) }^{ 2 }+{ (3-5) }^{ 2 }+{ (1-2) }^{ 2 }-{ (\cfrac { 2 }{ \sqrt { 3 } } ) }^{ 2 }\\ =6-\cfrac { 4 }{ 3 } \\ =\cfrac { 14 }{ 3 } \\ \therefore AD=\sqrt { \cfrac { 14 }{ 3 } } $$
Question 9
1 / -0

In which of these figures does the angle $$\theta$$ represent the inclination of line L.

A

B

C

D
None of the above

Solution

We know that inclination of a line is the angle it makes with the positive $$x-$$axis.
Hence, A will be correct.
Question 10
1 / -0

When the Y-coordinates of two points lying on a line are non-zero and equal, and the X-coordinates are unequal, then :

A
The line is Parallel to the X-axis.

B
The line is Parallel to the Y-axis.

C
The line is inclined at an angle of $$45^{\circ}$$ with the positive direction of the X-axis.

D
None of the above.

Solution

Equal y coordinates and unequal x coordinates means that the two points are equidistant from X axis and hence line containing them is parallel to x axis.
Question 11
1 / -0

Two adjacent sides of a rectangle are $$3{x^2} - 5{y^{2\,}}$$ and $$7{x^2} - xy$$. Find its perimeter?

A
$$10x²-10y² -2xy$$.

B
$$20x²-10y² -2xy$$.

C
$$x²-10y² -2xy$$.

D
$$40x²-10y² -2xy$$.

Solution

Adjacent sides in rectangle are lenght and base

Perimeter of rectangle =$$2(length + breadth)$$

Perimeter =$$2(3x²-5y²+7x²-xy)$$

Perimeter =$$2(10x² -5y²-xy)$$

Perimeter =$$20x²-10y² -2xy$$.
Question 12
1 / -0

What is the maximum positive value that slope of a line can take?

A
1.

B
$$\sqrt{2}$$

C
$$\sqrt{3}$$

D
$$\infty$$

Solution

Slope is tangent of its inclination angle. If angle is right angle, it's tangent tends to positive infinty.
Question 13
1 / -0

The point A $$(2a ,4a)$$, B$$(2a, 6a)$$ and C$$(2a + \sqrt 3 a, 5a)$$ (when $$a>0$$) are vertices of

A
an obtuse angled triangle

B
an equilateral triangle

C
an isosceles obtuse angled triangle

D
a right angled triangle

Solution

Distance AB = $$\sqrt {{{(2a - 2a)}^2} + {{\left( {6a - 4a} \right)}^2}} = \sqrt {4{a^2}} = 2a$$

Distance AC = $$\sqrt {{{(2a + \sqrt {3a} - 2a)}^2} + {{\left( {5a - 4a} \right)}^2}} $$

$$\sqrt {3{a^2} + {a^2}} = 2a$$

Distance BC= $$\sqrt {{{\left( {2a + \sqrt {3a} - 2a} \right)}^2} + {{\left( {5a - 6a} \right)}^2}} $$

$$\sqrt {3{a^2} + {a^2}} = 2a$$

AB = BC = CA

Equilateral Triangle
Question 14
1 / -0

If the projections of a line segment on the x, y and z axes in $$3$$-dimensional space are $$2, 3$$ and $$6$$ respectively, then the length of line segment is

A
$$6$$

B
$$7$$

C
$$9$$

D
$$12$$

Solution

Given projections are 2, 3 and 6

$$\therefore$$ AB $$=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$$

$$\Rightarrow AB=\sqrt{4+9+36}=\sqrt{49}=7$$.
Question 15
1 / -0

The distance between the points $$(3,5)$$ and $$(x,8)$$ is $$5$$ units. Then the value of $$x$$

A
$$7$$

B
$$2$$

C
$$5$$

D
none of these

Solution

Given the distance between the points $$(3,5)$$ and $$(x,8)$$ is $$5$$ units.
Then we have,
$$\sqrt{(x-3)^2+(8-5)^2}=5$$
or, $$(x-3)^2+3^2=5^2$$
or, $$(x-3)^2=5^2-3^2=4^2$$
or, $$x=\pm 4+3$$
or, $$x= 7,-1$$

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Straight Lines Test -4

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Straight Lines Test -4

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Question 1

( 3 , 0 )

( 0 , 3 )

( - 3 , 0 )

none of these.

Solution

Question 2

- 1

1

-2

none of these

Solution

Question 3

( - 3a , 2b )

( a , - 2b )

( 3a , - 2b)

none of these

Solution

Question 4

( 1, 2 )

( 1, 1 )

( - 1 , 2 )

none of these

Solution

Question 5

10

20

1/10

none of these

Solution

Question 6

$$-11$$

$$11$$

$$7$$

$$-7$$

Solution

Question 7

Only Statement I is True.

Only Statement II is True.

Both Statement I and Statement II are True.

Both Statement I and Statement II are False.

Solution

Question 8

$$2/\sqrt{3}$$

$$\sqrt{14/3}$$

$$16/\sqrt{3}$$

$$5/\sqrt{3}$$

Solution

Question 9

None of the above

Solution

Question 10

The line is Parallel to the X-axis.

The line is Parallel to the Y-axis.

The line is inclined at an angle of $$45^{\circ}$$ with the positive direction of the X-axis.

None of the above.

Solution

Question 11

$$10x²-10y² -2xy$$.

$$20x²-10y² -2xy$$.

$$x²-10y² -2xy$$.

$$40x²-10y² -2xy$$.

Solution

Question 12

1.

$$\sqrt{2}$$

$$\sqrt{3}$$

$$\infty$$

Solution

Question 13

an obtuse angled triangle

an equilateral triangle