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# Straight Lines

#### Multiple Choice Questions

1.

The equation of the straight line passing through the point (4, 3) and making intercepts on the co-ordinate axes whose sum is –1 is

D.

Let a and b be the intercepts on the co-ordinate axes. a + b = -1
⇒ b = - a - 1 = - (a + 1)

Equation of line is x/a + y/b = 1

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2.

The equation of bisectors of the angles between the lines  are

• y = 0 and x = 0

• None of the above

C.

y = 0 and x = 0

The equation of lines are ± x ± y = 0. Now, we take the lines x + y = 0 and x - y = 0.

$\therefore$  The equation of bisectors of the angles between these lines are

Hence, the equation of bisectors are x = 0 and y = 0.

3.

A body falling from rest under gravity passes a certain point P. It was at a distance of 400 m from P, 4s prior to passing through P. If g = 10 m/s2 , then the height above the point P from where the body began to fall is

• 720 m

• 900 m

• 320 m

• 320 m

A.

720 m

Subtracting we get 400 = 8g + 4gt
⇒ t = 8 sec
∴
∴ Desired height = 320 + 400 = 720 m.
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4.

If one of the lines given by 6x2 -xy +4cy2 = 0 is 3x + 4y = 0, then c equals

• 1

• -1

• 3

• 3

D.

3

The pair of lines is 6x2-xy +4cy2 =0
On comparing with ax2 +2hxy by2 = 0
we get a = 6, 2h =-1, b= c
therefore
m1 +m2 = -2h/b = 1/4c and m1m2 = a/b = 6/4c
On line of given pair of lines is
3x+4y = 0
slope of line = -3/4 = m1
-3/4 +m2 = 1/4c
m2 = 1/4c + 3/4

1 + 3c = -8 3c = -9
⇒ c = -3

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5.

Two points A and B move from rest along a straight line with constant acceleration f and f′ respectively. If A takes m sec. more than B and describes ‘n’ units more than B in acquiring the same speed then

• (f - f′)m2 = ff′n

• (f + f′)m2 = ff′n

• 1/2(f - f′)m = ff′n2

• 1/2(f - f′)m = ff′n2

D.

1/2(f - f′)m = ff′n2

v2 = 2f(d + n) = 2f′d
v = f′(t) = (m + t)f
eliminate d and m we get
(f'-f)n =1/2ff'm2

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6.

A straight line through the point A(3, 4) is such that its intercept between the axes is bisected at A. Its equation is

• x + y = 7

• 3x − 4y + 7 = 0

• 4x + 3y = 24

• 4x + 3y = 24

C.

4x + 3y = 24

The equation of axes is xy = 0
⇒ the equation of the line is

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7.

The intercept on the line y = x by the circle x2 +y2 -2x = 0 is AB. Equation of the circle on AB as a diameter is

• x2 +y2 -x-y =0

• x2 -y2 -x-y =0

• x2 +y2 +x-y =0

• x2 +y2 +x-y =0

A.

x2 +y2 -x-y =0

The equation of line is y = x ........... (i)
and equation of circle is x2 + y2 - 2x = 0 ........... (ii)
On solving equation (i) and equation (ii),
we get x2 + x2 - 2x = 0
2x2 - 2x = 0
= 2x(x - 1) = 0
x = 0, x = 1
when x = 0, y = 0
when x = 1, y = 1
Let coordinate of A is (0, 0) and co-ordinate of B is (1, 1)

∴ Equation of circle (AB as a diameter)
(x - x1 )(x - x2 ) + (y - y1 )(y - y2 ) = 0
(x - 0)(x - 1) + (y - 0)(y - 1) = 0
x(x - 1) + y(y - 1) = 0
x2 - x + y2 - y = 0
x2 + y2 - x - y = 0

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8.

The system of equations
αx + y + z = α - 1,
x + αy + z = α - 1,
x + y + αz = α - 1

has no solution, if α is

• -2

• either-2 or 1

• not -2

• not -2

A.

-2

αx + y + z = α - 1,
x + αy + z = α - 1,
x + y + αz = α - 1

= α(α2 – 1) – 1(α - 1) + 1(1 - α)
= α (α - 1) (α + 1) – 1(α - 1) – 1(α - 1)
⇒ (α - 1)[α2 + α - 1 – 1] = 0
⇒ (α - 1)[α2 + α - 2]= 0 [α2 + 2α - α - 2] = 0
(α - 1) [α(α + 2) – 1(α + 2)]= 0
(α - 1) = 0,
α + 2 = 0 ⇒ α = –2, 1; but α ≠ 1

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9.

The two lines x = ay + b, z = cy + d; and x = a′y + b′, z = c′y + d′ are perpendicular to each other if

• aa′ + cc′ = −1

• aa′ + cc′ = 1

A.

aa′ + cc′ = −1

Equation of lines

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10.

If |z + 4| ≤ 3, then the maximum value of |z + 1| is

• 4

• 10

• 6

• 6

C.

6

114 Views