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
The equation of bisectors of the angles between the lines $\left|\mathrm{x}\right|=\left|\mathrm{y}\right|$ are
$\mathrm{y}=\pm \mathrm{x}\mathrm{and}\mathrm{x}=0$
$\mathrm{x}=\frac{1}{2}\mathrm{and}\mathrm{y}=\frac{1}{2}$
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
$\frac{\mathrm{x}+\hspace{0.17em}\mathrm{y}}{\sqrt{1+1}}=\pm \frac{{\displaystyle \mathrm{x}-\hspace{0.17em}\mathrm{y}}}{{\displaystyle \sqrt{1+1}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{x}+\hspace{0.17em}\mathrm{y}=\pm \left(\mathrm{x}-\mathrm{y}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Taking}+\mathrm{ve}\mathrm{sign},\mathrm{x}+\mathrm{y}=\mathrm{x}-\mathrm{y}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{y}=0\phantom{\rule{0ex}{0ex}}\mathrm{Taking}-\mathrm{ve}\mathrm{sign},\mathrm{x}+\mathrm{y}=-\left(\mathrm{x}-\mathrm{y}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{x}=0$
Hence, the equation of bisectors are x = 0 and y = 0.
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/s^{2} , 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
If one of the lines given by 6x^{2} -xy +4cy^{2} = 0 is 3x + 4y = 0, then c equals
1
-1
3
3
D.
3
The pair of lines is 6x^{2}-xy +4cy^{2} =0
On comparing with ax^{2} +2hxy by^{2} = 0
we get a = 6, 2h =-1, b= c
therefore
m_{1} +m_{2} = -2h/b = 1/4c and m_{1}m_{2} = a/b = 6/4c
On line of given pair of lines is
3x+4y = 0
slope of line = -3/4 = m1
-3/4 +m_{2} = 1/4c
m_{2} = 1/4c + 3/4
1 + 3c = -8 3c = -9
⇒ c = -3
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′)m^{2} = ff′n
1/2(f - f′)m = ff′n^{2}
1/2(f - f′)m = ff′n^{2}
D.
1/2(f - f′)m = ff′n^{2}
v2 = 2f(d + n) = 2f′d
v = f′(t) = (m + t)f
eliminate d and m we get
(f'-f)n =1/2ff'm^{2}
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
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
x^{2} +y^{2} -x-y =0
x^{2} -y^{2} -x-y =0
x^{2} +y^{2} +x-y =0
x^{2} +y^{2} +x-y =0
A.
x^{2} +y^{2} -x-y =0
The equation of line is y = x ........... (i)
and equation of circle is x^{2} + y^{2} - 2x = 0 ........... (ii)
On solving equation (i) and equation (ii),
we get x^{2} + x^{2} - 2x = 0
2x^{2} - 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 - x_{1} )(x - x_{2} ) + (y - y_{1} )(y - y_{2} ) = 0
(x - 0)(x - 1) + (y - 0)(y - 1) = 0
x(x - 1) + y(y - 1) = 0
x^{2} - x + y^{2} - y = 0
x^{2} + y^{2} - x - y = 0
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
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