Straight Lines - Online Test

Q1. The lines x + ( k – 1) y + 1 = 0 and are at right angles if
Answer : Option A
Explaination / Solution:

If the lines are at right angles to each other, then the product of their slopes = -1. Slope of any line = -(coefficient of x/coeffiecient of y)

Therefore slope of line 1 = -(1/(k-1)

Slope of line 2 = -2/k2

Therefore  X  = - 1

That is k2 (k-1) = -2

i.e; k3 - k2 +2 = 0

On factorizing we get (k+1)(k2 -2k - 2) = 0

This imlpies k+1 is a factor, hence k = -1

Hence they are at right angles if k = -1


Q2. A line is equally inclined to the axis and the length of perpendicular from the origin upon the line is √2. A possible equation of the line is
Answer : Option B
Explaination / Solution:

Since the line is equally inclined the slope of the line should be -1, because it makes 1350 in the positive direction of the X axis

This implies the equation of the line is y= -x +c 

i.e; x+y - c =0

distance of the line from the origin is given as 

Therefore  = 

This implies c = 2

Hence the equation of the line is x+y=2


Q3. The lines y = mx , y + 2x = 0 , y = 2x + λ and y = - mx + λ form a rhombus if m =
Answer : Option B
Explaination / 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+  and y = -mx +  are parallel.

If two lines are equal, then their slopes are qual

This implies  m = -2


Q4. Two points ( a , 0 ) and ( 0 , b ) are joined by a straight line. Another point on this line is
Answer : Option A
Explaination / 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.


Q5. The coordinates of the foot of perpendicular from ( 0 , 0 ) upon the line x + y = 2 are
Answer : Option C
Explaination / 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)


Q6. The area of the triangle whose sides are along the lines x = 0 , y = 0 and 4x + 5y = 20 is
Answer : Option B
Explaination / Solution:

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


Q7. A line is drawn through the points ( 3 , 4 ) and ( 5 , 6 ) . If the is extended to a point whose ordinate is – 1, then the abscissa of that point is
Answer : Option D
Explaination / Solution:

The slope of the given line is  =  = 1

Therefore  = 1

That is 4+1 = 3 - x

Therefore x = -2


Q8. The line x + y – 6 = 0 is the right bisector of the segment [PQ]. If P is the point ( 4, 3 ) , then the point Q is
Answer : Option B
Explaination / Solution:

Equation of the line which is perpendicular to the given line x + y = 6 is x - y + k = 0

Since it passes through the point (4,3)

4 - 3 + k = 0

Therefore k = -1

Hence the equation of the line segment PQ is x - y - 1 = 0

On solving these two lines we get x = 7/2 and y = 5/2

This point of intersection is the midpoint of the line segement PQ

That is  = 7/2 Hence x = 3

Similarly  = 5/2. Hence y = 2

Hence the coordiates of the point Q is (3,2)


Q9. The acute angle between the lines ax + by + c = 0 and ( a + b )x = ( a – b )y , a ≠ b , is
Answer : Option C
Explaination / Solution:

The slope of the first line is -(a/b) and slope of the second line is 

Angle between the lines is   = 

Let m1 =-(a/b) and m2 = 

Hence  =[ 

On simplifying we get, 

 = 1

This implies  = 450


Q10. The lines x + 2y – 3 = 0, 2x + y – 3 = 0 and the line l are concurrent . If the line I passes through the origin, then its equation is
Answer : Option C
Explaination / Solution:

Equation of a line passing through the intersection of two lies is given by ax1 +by1 +c1 + k(ax2 +by2 +c2) = 0

Hence x+2y-3 + k(2x+y-3) = 0

Since it passes through (0,0)

-3 -3k = 0

This implies k = -1

Sustituting for k we get,

x+2y-3 +(-1)(2x+y-3) = 0

-x +y =0 or x - y = 0