# Straight Lines - Online Test

Q1. The equation of the line which passes through the point ( 1 , - 2 ) and cuts off equal intercepts from the axis is
Explaination / Solution:

Let the equation of the line which has equal intercepts be

+  =1

Substituting the values for x and y

1 -2 =a

Therefore a = -1

Hence the equation of the line is x+y+1=0

Q2. If a line is drawn through the origin and parallel to the line x – 2y + 5 = 0 , its equation is
Explaination / Solution:

If the line is parallel to the given line x-2y + 5 =0,

then the required line will have same slope

Hence the equaion of the given line is x-2y+k=0

since it passes through the origin,

0+0+k=0

Therefore k=0

Hence the equation of the required line is x-2y=0 or x=2y

Q3. The line passing through ( 1, 1 ) and parallel to the line 2x – 3y + 5 = 0 is
Explaination / Solution:

The required line is parallel to the given line, hence it has same slope

Therefore the equation of the required line is 2x-3y+k=0

Since it passes through (1,1)

2(1) - 3(1) +k = 0

Therefore k=1

Hence the equation of the required line is 2x - 3y + 1=0

Q4. The equation of the line which passes through the point ( 2 , - 3 ) and cuts off equal intercepts from the axis is
Explaination / Solution:

The equation of the line which cuts equal intercepts is

x+y=a

Since it passes through (2,-3),

2 - 3 =a

Hence a = -1

Hence the equation of the required line is x + y +1 = 0

Q5. The straight lines x + y - 4 = 0 , 3x + y – 4 = 0 , x - 3y – 4 = 0 form a triangle which is
Explaination / Solution:

The triangle formed by these lines is a right angled triangle

If the lines are perpendicular to each other, then the product of their slopes is -1

The slope of lines  3x + y – 4 = 0 , x - 3y – 4 = 0 are  -3 and 1/3 respectively.

The product of the slopes is -1

Hence these two lines are perpendicular to each other

This infers that the triangle formed by these lines is a right angled triangle.

Q6. The distance between the parallel lines 4x + 3y + 11 = 0 and 8x + 6y = 15 is
Explaination / Solution:

Distance between two parallel lines is given by

c1 = 2 x 11 = 22 and c2 = 15 and A = 8 and B = 6

Now substituting the values we get

= 7/10

Q7.

The line which passes through the point ( 0 , 1 ) and perpendicular to the line x – 2y + 11 = 0 is

Explaination / Solution:

The line which is perpendicular to the given line is 2x + y + k = 0

Since it passes through (0,1)

2(0) + 1 + k = 0

This implies k = -1

Hence the equation of the required line is 2x + y  - 1 = 0

Q8. The perpendicular distance of the origin from the line 3x +4y + 1 = 0 is
Explaination / Solution:

The perpendicular distance from the origin to the line is given by

For the given line c = 1, A = 3 and B = 4 and since it passes through the origin,

Sustituting the values we get,

= 1/5

Q9. The acute angle between the lines x – y = 0 and y = 0 is
Explaination / Solution:

Slope of the line x - y = 0 is 1. This imlpies the tan = 1 Hence the angle made by the line with X- axis is 450

The angle made by the line y = 0 with the Y axis is 900

Therefore the acute angle between the lines is 450

Q10. The vertices of a triangle are ( 0 , 3 ) , ( - 3 , 0 ) and ( 3 , 0 ) . The orthocenter of the triangle is
Explaination / 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)