Solved Objective Question on Straight line Set 2

Q1.

The straight line y = x–2 rotates about a point where it cuts the x-axis and becomes perpendicular to the straight line ax + by + c = 0. Then its equation is.

(A) ax + by + 2a = 0

(B) ax – by – 2a = 0

(D) ay – bx + 2b = 0

Solution:

Slope of the line in the new position is b/a, since it is perpendicular to the line ax + by + c = 0 and it cuts the x-axis. Hence the required line passes through (2, 0) and its slope is b/a.

The required equation is y – 0 = b/a (x-2)

or, ay = bx – 2b or, ay – bx + 2b = 0.

Hence (D) is the correct answer.

Q2.

Consider the equation y - y₁ = m(x-x₁). If m and x₁ are fixed and different lines are drawn for different values of y₁, then.

(A) the line will pass through a fixed point.

(B) there will be a set of parallel lines.

(D) all the lines will be parallel to the line y=y₁.

Solution:

For a fixed value of m, the given lines form a set of parallel lines all with the slope m,
&⇒ answer (B) is correct, while (A) is incorrect. However this set will not be parallel to the line y = y₁, whose slope is zero, unless
m = 0. And (D) is therefore excluded. If, furthermore, x₁ is fixed, the above set of lines will all intersect the line x = x₁. Hence answer (C) is also correct. Hence (B) and (C) are the correct answers.

Q3.

Let 2x–3y =0 be a given line and P (sinq, 0) and Q (0, cosq) be the two points. Then P and Q lie on the same side of the given line, if q lies in the .

(A) 1st quadrant

(B) 2nd quadrant

(D) 4th quadrant

Solution:

P and Q lie on the same side if 2sinq and - 3cosq have the same signs i.e. sinq and cosq have opposite signs which is true for the 2nd and 4th quadrant. .

Hence (B) and (D) are the correct answers.

Q4.

The straight lines of the family x(a + b) + y (a – b) = 2a (a and b being parameters) are

(A) not concurrent

(B) Concurrent at (1, -1)

(D) None of these

Solution:

The given equation can be written as

a(x + y – 2) + b(x – y) = 0 or (x + y – 2) + b/a(x – y) = 0

This is a family of lines concurrent at x – y = 0 and x + y – 2 = 0

On solving these two equations we get (1, 1)

Hence (C) is the correct answer.

Q5.

Drawn from origin are two mutually perpendicular lines forming an isosceles triangle together with the straight line 2x+y = a. Then the area of this triangle is.

(A)

(B)

(C)

(D) None of these

Solution:

Let the two perpendiculars through the origin intersect 2x+y = a at A and B so that the triangle OAB is isosceles.

OM = length of perpendicular from

O to AB, OM = .Also AM = MB = OM

&⇒ AB = . Area of D OAB = =

Hence (C) is the correct answer. .

Q6.

A curve with equation of the form y = ax⁴ + bx³+cx +d has zero gradient at the point (0, 1) and also touches the x-axis at the point ( -1, 0). Then the values of x for which the curve has negative gradients are.

(A) x > -1

(B) x < 1

(D) -1 £ x £ 1

Solution:

y = ax⁴ + bx³ +cx +d . . . . (1).

y touches x-axis at (-1, 0)

so, ( -1, 0) lies on it and dy/ dx =0

so, 0 = a – b – c +d . . . . (2).

From (1) dy/ dx = 4ax³ + 3bx² +c . . . . (3).

Hence
&⇒ - 4a + 3b +c =0 . . . . (4)

Also = c =0 (since curve touches (0, 1) )

&⇒ ( 0, 1) also lies on it. Hence d = 1.

Putting values of c and d in (2) and (4), solving for a and b we get a = 3, b = 4. Therefore (3) becomes = 12x³ +12x²

Now dy/dx < 0
&⇒ 12 x³ + 12x²< 0
&⇒ 12x² (x+1) < 0
&⇒ x < -1

Hence (C) is the correct answer.

Q7.

The area of triangle is 5. Two of its vertices are (2, 1) and (3, -2), the third vertex is lying on y = x + 3. The co-ordinates of the third vertex can be.

(a)

(b)

(c)

(d)

Solution:

As the third vertex lies on the line y = x+3, its co-ordinates are of the form (x, x+3). .

The area of triangle is |4x-4| = |2x-2|

According to given condition
&⇒ 2x-2 = ± 5
&⇒ x = -3/2, 7/2

Hence the coordinates of third vertex can be

Hence (A) and (C) are the correct answers.

Q8.

The orthocentre of the triangle formed by the lines 2x² + 3xy – 2y² – 9x + 7y – 5 = 0, 4x + 5y – 3 = 0 lies at

(A) (3/5 , 11/5)

(B) (6/5, 11/5)

(D) None of these

Solution:

The pair of straight lines 2x² + 3xy – 2y² – 9 x + 7y– 5 = 0 are perpendicular to each other so orthocentre is point of intersection of these lines. Hence (A) is correct.