General equation of second degree
The general equation of first degree in x and y always represents a straight line.
Let A_{1}x + B_{1}y + C_{1} = 0.......................(i)
and A_{2}x + B_{2}y + C_{2} = 0.......................(ii)
be the equations of two straight lines.
Now, Combining these equations we get,
(A_{1}x + B_{1}y + C_{1}) (A_{2}x + B_{2}y + C_{2}) = 0........................(iii)
The coordinates of any point, which satisfy the equation (iii), will also satisfy either equation (i) or equation (ii).
Similarly, the co-ordinates of any point, which satisfy any one of the equation (i) or (ii) will also satisfy the equation (iii).
Therefore, equation (iii) represents two separate straight lines (i) and (ii). In other words, equation (iii) represents a pair of the straight lines given by (i) and (ii). So, (iii) is the equation of a pair of lines.
Now, Expanding the left hand side of equation (iii) we get,
A_{1}A_{2}x^{2} + (A_{1}B_{2} + A_{2}B_{1})xy + B_{1}B_{2}y^{2} + (A_{1}C_{2} + A_{2}C_{1})x + (B_{1}C_{2} + B_{2}C_{1})y + C_{1}C_{2} = 0
If we put A_{1}A_{2} = a,A_{1}B_{2} + A_{2}B_{1} = 2h, B_{1}B_{2} = b, A_{1}C_{2} + A_{2}C_{1} = 2g, B_{1}C_{2} + B_{2}C_{1 }= 2f and C_{1}C_{2} = c, then the above equation becomes,
ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0..........................(iv)
This equation is called the general equation of the second degree in x and y. Thus we see that the equation of a pair of lines is a second degree equation. But the converse of this statement is not always true. It means every second degree equations in x and y may not represent a pair of straight lines. Equations of second degree will represent a pair of straight lines only if the left hand side can be resolved into two linear factors.
Consider an equation y^{2} - 3xy + 2x^{2} = 0. This equation is equivalent to (y - x) (y - 2x) = 0.
So, the equation y^{2} - 3xy + 2x^{2} = 0 represents the two straight lines y - x = 0 and y - 2x = 0.
Similarly, the equation xy = 0 represents the two straight lines x = 0 and y = 0.
Again,
Consider an equation x^{2} - 5x + 6 = 0 represents two straight lines x - 2 = 0 and x - 3 = 0.
And the equation x^{2} - y^{2} = 0 represents the two straight lines x + y = 0 and x - y = 0.
Condition that the general equation of second degree may represent a line pair
The general equation of second degree in x and y is
ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0
or, ax^{2} + (2hy + 2g)x + (by^{2} + 2fy + c) = 0
This is quadratic equation in x.
So, x = \(\frac {-(2hy + 2g) ± \sqrt {(2hy + 2g)^2 - 4a(by^2 + 2fy + c)}}{2a}\)
or, x = \(\frac {-(hy + g) ± \sqrt {(hy + g)^2 - a(by^2 + 2fy + c)}}{a}\)
These two equations will be linear if
(hy + g)^{2} - a (by^{2} + 2fy + c) is a perfect square.
or, h^{2}y^{2} + 2ghy + g^{2} - aby^{2} - 2afy - ac is a perfect square.
i.e. (h^{2} - ab) y^{2} + (2gh - 2af) y + (g^{2} - ac) is a perfect square.
i.e. (2gh - 2af)^{2} - 4 (h^{2} - ab) (g^{2} - ac) = 0
i.e. g^{2}h^{2} - 2ghaf + a^{2}f^{2} - g^{2}h^{2} + h^{2}ac + abg^{2} - a^{2}bc = 0
i.e. a (af^{2} + bg^{2} + ch^{2} - 2fgh - abc) = 0
i.e. abc + 2fgh - af^{2} - bg^{2} - ch^{2} = 0
Hence, the general equation of second degree ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 may represent a line pair if abc + 2fgh - af^{2} - bg^{2} - ch^{2} = 0.
Quadratic Equation
Any equation in the form of ax^{2} + bx + c = 0 is called quadratic equation in x.
Multiplying both sides of this equation by 4a we get,
4a (ax^{2} + bx + c) = 0
or, 4a^{2}x^{2} + 4abx + 4ac = 0
or, (2ax)^{2} + 2 . 2ax . b + b^{2} - b^{2} + 4ac = 0
or, (2ax + b)^{2} = b^{2} - 4ac
or, (2ax + b)^{2} = (\(\sqrt {b^2 - 4ac}\))^{2}
or, 2ax + b = ± \(\sqrt {b^2 - 4ac}\)
or, 2ax = - b ± \(\sqrt {b^2 - 4ac}\)
or, x = \(\frac {- b ± \sqrt {b^2 - 4ac}}{2a}\)
Let α = \(\frac {- b + \sqrt {b^2 - 4ac}}{2a}\) and β = \(\frac {- b - \sqrt {b^2 - 4ac}}{2a}\).
Then α and β are called roots of the quadratic equation ax^{2} + bx + c = 0.
Now,
\begin{align*} \text{Sum of the roots (α + β)} &= \frac {- b + \sqrt {b^2 - 4ac}}{2a} + \frac {- b - \sqrt {b^2 - 4ac}}{2a}\\ &= - \frac ba\\ &= -\frac {coefficient\;of\;x}{coefficient\;of\;x^2}\\ \end{align*}
\begin{align*} \text{Sum of the roots (αβ)} &= (\frac {- b + \sqrt {b^2 - 4ac}}{2a}) (\frac {- b - \sqrt {b^2 - 4ac}}{2a})\\ &= \frac ca\\ &= \frac {constant\;term}{coefficient\;of\;x^2}\\ \end{align*}
Homogeneous equation of second degree
An equation in x and y in which the sum of the power of x and y in every term is the same, is called homogenous equation. If this sum is two, then the equation is called a homogenous equation of second degree. The equation ax^{2} + 2hxy + by^{2} = 0 is the general homogenous equation of the second degree.
A homogeneous equation of the second degree represents a pair of straight lines which pass through the origin.
Proof: Consider the homogenous equation of the second degree.
ax^{2 }+ 2hxy + by^{2} = 0...........................(i)
or, by^{2} + 2hxy + ax^{2} = 0
If b ≠ 0, the equation can be written as y^{2} + \(\frac {2h}b\)xy + \(\frac ab\)x^{2} = 0
or, (\(\frac yx\))^{2} + \(\frac {2h}b\)(\(\frac yx\)) + \(\frac ab\) = 0
This is quadratic equation in \(\frac yx\). So it has two roots. Let these roots be m_{1} and m_{2}.
Then,
\(\frac yx\) = m_{1} and \(\frac yx\) = m_{2}
or, y = m_{1}x and y = m_{2}x.
These two equations are the equations of straight lines passing through the origin.
If b = 0, then equation (i) becomes
ax^{2} + 2hxy = 0
or, x(ax + 2hy) = 0 which represents two straight lines x = 0 and ax + 2hy = 0.
These two lines pass through the origin.
Hence, the homogenous equation of second degree always represents a pair of straight lines passing through the origin.
Angle between the line pair represented by ax^{2} + 2hxy + by^{2} = 0
Homogeneous equation of second degree is:
ax^{2} + 2hxy + by^{2} = 0
or, by^{2} + 2hxy + ax^{2} = 0
or, y^{2} + \(\frac{2h}{b}\)(\(\frac yx\)) + \(\frac ab\) = 0.........................(i)
This is quadratic equation in \(\frac yx\). So, it has two roots. Let these two roots be m_{1} and m_{2}.
Then,
\(\frac yx\) = m_{1} and \(\frac yx\) = m_{2}
or, y = m_{1}x and y = m_{2}x which are two seperate equations represente by the given equation ax^{2} + 2hxy + by^{2} = 0.
Now,
From the quadratic equation (i)
m_{1} + m_{2} = \(\frac {-2h}b\) and m_{1}m_{2} = \(\frac ab\)
Now,
\begin{align*} m_1- m_2 &= \sqrt {(m_1 + m_2)^2 - 4m_1m_2}\\ &= \sqrt {\frac{4h^2}{b^2} - 4\frac ab}\\ &= \sqrt {\frac {4h^2 - 4ab}{b^2}}\\ &= \frac 2b \sqrt {h^2 - ab}\\ \end{align*}
Let \(\theta\) be the angle between the lines y = m_{1}x and y = m_{2}x. Then,
\begin{align*} tan\theta &= ± \frac {m_1 - m_2}{1 + m_1m_2}\\ &= ± \frac {\frac 2b \sqrt {h^2 - ab}}{1 + \frac ab}\\ &= ± \frac {2\sqrt {h^2 - ab}}{a + b}\\ \end{align*}
∴ \(\theta\) = tan^{-1 }(± \(\frac {2\sqrt {h^2 - ab}}{a + b}\))
Second Method:
We have ax^{2} + 2hxy + by^{2} = 0............................................(i)
Let the seperate equations represented by this equation be y = m_{1}x and y = m_{2}x.
Now,
The combined equation of theses equation is:
(y - m_{1}x) (y - m_{2}x) = 0
or, y^{2} - (m_{1} + m_{2})xy + m_{1}m_{2}x^{2} = 0
or, m_{1}m_{2}x^{2}- (m_{1} + m_{2})xy + y^{2} = 0.......................................(ii)
Comparing (i) and (ii) we have,
\(\frac {m_1m_2}{a}\) = \(\frac {- (m_1 + m_2)}{2h}\) = \(\frac 1b\)
Taking 1^{st} and last, m_{1}m_{2} = \(\frac ab\)
Taking 2^{nd} and last, m_{1} + m_{2} = -\(\frac {2h}b\)
Now,
\begin{align*} m_1 - m_2 &= \sqrt {(m_1 + m_2)^2 - 4m_1m_2}\\ &= \sqrt {\frac {4h^2}{b^2} - 4\frac ab}\\ &= \sqrt {\frac {4h^2 - 4ab}{b^2}}\\ &= \frac 2b \sqrt {h^2 - ab}\\ \end{align*}
Let \(\theta\) be the angle between the lines y = m_{1}x and y = m_{2}x.
Then,
\begin{align*} tan\theta &=± \frac {m_1 - m_2}{1 + m_1m_2}\\ &= ± \frac {\frac 2b \sqrt {h^2 - ab}}{1 + \frac ab}\\ &= ± \frac {2\sqrt {h^2 - ab}}{a + b}\\ \end{align*}
∴ \(\theta\) = tan^{-1} (± \(\frac {2\sqrt {h^2 - ab}}{a + b}\))
Condition that the straight lines given by the equation ax^{2} + 2hxy + by^{2} = 0 may be (1) Perpendicular and (2) Coincident.
If the equation ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 represents a pair of lines, then ax^{2} + 2hxy + by^{2} = 0 represents a pair of lines through the origin parallel to the above pair.
Proof:
Let ax^{2}+ 2hxy + by^{2} + 2gx + 2fy + c = 0............................................................(i)
represent a pair of straight lines. Then the left-hand side can be resolved into two linear factors. Let these factors be A_{1}x + B_{1}y + C_{1}and A_{2}x + B_{2}y + C_{2} = 0.
Now
Combining equation of these equations is:
(A_{1}x + B_{1}y + C_{1}) (A_{2}x + B_{2}y + C_{2}) = 0
or, A_{1}A_{2}x^{2} + (A_{1}B_{2} + A_{2}B_{1})xy + B_{1}B_{2}y^{2} + (A_{1}C_{2} + A_{2}C_{1})x + (B_{1}C_{2} + B_{2}C_{1})y + C_{1}C_{2} = 0..............................................(ii)
Equating the coefficients of like terms in equation (i) and (ii) we have,
A_{1}A_{2} = a^{2}, A_{1}B_{2} + A_{2}B_{1} = 2h, B_{1}B_{2} = b^{2}, A_{1}C_{2} + A_{2}C_{1} = 2g, B_{1}C_{2} + B_{2}C_{1} = 2f, C_{1}C_{2} = c^{2}
Now,
Equation of the straight line parallel to A_{1}x + B_{1}y + C_{1} = 0 and passing through the origin is:
A_{1}x+ B_{1}y = 0.........................................(iii)
Again,
Equation of straight line parallel to A_{2}x + B_{2}y + C_{2} = 0 and passing through the origin is:
A_{2}x + B_{2}y = 0.........................................(iv)
Now,
Combining (iii) and (iv) we have,
(A_{1}x + B_{1}y) (A_{2}x + B_{2}y) = 0
or, A_{1}A_{2}x^{2} + (A_{1}B_{2} + A_{2}B_{1})xy + B_{1}B_{2}y^{2} = 0
or, ax^{2} + 2hxy + by^{2} = 0. This completes the proof.
Note:Angles between the line pair ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 are same as the angles between the pair ax^{2} + 2hxy + by^{2} = 0.
The angles are given by,
tan^{-1} (± \(\frac {2\sqrt {h^2 - ab}}{a + b}\))
Hence, the two lines will be perpendicular to each other if a + b = 0 and they will beparallel if h^{2}= ab.
General equation of the second degree :
ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0.
.
Here,
3x^{2} - 5xy - 2y^{2} - x + 2y = 0
or, 3x^{2} - 6xy + xy - 2y^{2} - x + 2y = 0
or, 3x(x - 2y) + y(x - 2y) - 1(x - 2y) = 0
or, (x - 2y) (3x + y - 1) = 0
Either: x - 2y = 0
Or: 3x + y - 1 = 0
∴ The required equations are: x - 2y = 0 and 3x + y - 1 = 0_{Ans}
Here,
x^{2} - 3xy + 2y^{2} = 0
or, x^{2} - 2xy - xy + 2y^{2} = 0
or, x(x - 2y) - y(x - 2y) = 0
or, (x - 2y) (x - y) = 0
Either: x - 2y = 0
Or: x - y = 0
∴ The required seperate equations are:x - 2y = 0 andx - y = 0 _{Ans}
Here,
The given equation is:6x^{2} + 11xy - 6y^{2} = 0..................(1)
The homogeneous equation of second is: ax^{2} + 2hxy + by^{2} = 0.................(2)
Comparing (1) and (2)
a = 6
2h = 11 i.e. h = \(\frac {11}2\)
b = -6
Now.
a + b = 0
or, 6 - 6 = 0
∴ 0 = 0
Hence, a + b = 0 is satisfied by the given equation so the equation6x^{2} + 11xy - 6y^{2} = 0 are perpendicular to each other. _{Proved}
Here,
Given equation is:x^{2} - 4xy + 4y^{2} = 0......................(1)
The homogenous equation of second degree is:
ax^{2}+ 2hxy + by^{2} = 0...........................(2)
Comparing (1) and (2)
a = 1
2h = -4 i.e. h = \(\frac {-4}2\) = -2
b = 4
Now,
h^{2} = ab
or, (-2)^{2} = 1× 4
∴ 4 = 4
Hence, h^{2} = ab is satisfied by the given equation so the equationx^{2} - 4xy + 4y^{2} = 0 are coincident to each other. _{Proved}
Here,
ax^{2} + 2hxy + by^{2} = 0
When: h^{2} = ab then straight lines are coincident each other.
When: a + b = 0 then straight lines are perpendicular each other.
Given eq^{n} of line are:
x = 2y
i.e. x - 2y = 0..........................(1)
2x = y
i.e. 2x - y = 0..........................(2)
Combined eq^{n} of (1) and (2) is:
(x - 2y) (2x - y) = 0
or, 2x^{2}- xy - 4xy + 2y^{2} = 0
or, 2x^{2} - 5xy + 2y^{2} = 0
∴ The required eq^{n} is: 2x^{2} - 5xy + 2y^{2} = 0 _{Ans}
Here,
Given eq^{n} is: y^{2} = x^{2}
or, x^{2} - y^{2} = 0
or, (x - y) (x + y) = 0
either: x + y = 0
Or, x - y = 0
Slope of eq^{n} x + y = 0 is: m_{1} = -\(\frac 11\) = -1
Slope of eq^{n} x -y = 0 is: m_{2} = -\(\frac 1{-1}\) = 1
m_{1}× m_{2} = -1× 1 = -1
Hence, they are perpendicular to each other. _{Proved}
Here,
Given eq^{n} is:
6x^{2} - 5xy - 6y^{2} = 0...............................(1)
The eq^{n}of homogenous is:
ax^{2} + 2hxy + by^{2} = 0..........................(2)
Comparing eq^{n} (1) and (2)
a = 6
b = -6
Now,
a + b = 0
or, 6 - 6 = 0
∴ 0 = 0
Hence, the angle between two lines is 90°. _{Proved}
Here,
Given eq^{n} is:
x^{2} - 2xy sec\(\alpha\) + y^{2} = 0...........................(1)
The homogenous eq^{n} is:
ax^{2} + 2hxy + by^{2} = 0..............................(2)
Comparing eq^{n} (1) and (2)
a = 1
h = - sec\(\alpha\)
b = 1
Now,
tan\(\theta\) =± \(\frac {2\sqrt {h^2 - ab}}{a + b}\)
or, tan\(\theta\) =± \(\frac {2\sqrt {(-sec\alpha)^2 - 1 × 1}}{1 + 1}\)
or, tan\(\theta\) =± \(\frac {2\sqrt {sec^2\alpha - 1}}2\)
or, tan\(\theta\) =± \(\sqrt {sec^2\alpha - 1}\)
or, tan\(\theta\) =± \(\sqrt {tan^2\alpha}\)
or, tan\(\theta\) =± tan\(\alpha\)
∴ \(\theta\) = \(\alpha\) _{Ans}
Here,
Given eq^{n} are:
x cos\(\alpha\) + y sin\(\alpha\) = 0...................(1)
x sin\(\alpha\) + y cos\(\alpha\) = 0...................(2)
Combined eq^{n} of (1) and (2)
(x cos\(\alpha\) + y sin\(\alpha\)) (x sin\(\alpha\) + y cos\(\alpha\)) = 0
or, x^{2} cos\(\alpha\).sin\(\alpha\) +xy cos^{2}\(\alpha\) + xy sin^{2}\(\alpha\) + y^{2} sin\(\alpha\).cos\(\alpha\) = 0
or, x^{2} cos\(\alpha\).sin\(\alpha\) + y^{2} sin\(\alpha\).cos\(\alpha\) + xy (cos^{2}\(\alpha\) + sin^{2}\(\alpha\)) = 0
or, sin\(\alpha\).cos\(\alpha\) (x^{2} + y^{2}) + xy× 1 = 0
∴(x^{2} + y^{2}) sin\(\alpha\).cos\(\alpha\) + xy = 0 _{Ans}
Given equation is:
x^{2} + 4xy + y^{2} = 0.............................(1)
The homogeneous equation of second degree is:
ax^{2} + 2hxy + by^{2} = 0.....................(2)
Comparing (1) and (2)
a = 1
h =2
b = 1
Now,
tan\(\theta\) = ± \(\frac {2\sqrt {h^2 - ab}}{a + b}\)
or, tan\(\theta\) = ± \(\frac {2\sqrt {2^2 - 1 × 1}}{1 + 1}\)
or, tan\(\theta\) = ± \(\frac {2\sqrt {4 - 1}}2\)
or, tan\(\theta\) = ± \(\sqrt 3\)
∴ obtuse angle (\(\theta\) = tan^{-1} (-\(\sqrt 3\))
∴ \(\theta\) = (90 + 30)° = 120°_{Ans}
Here,
Given equation is:
3x^{2} + 2y^{2} - 5xy = 0
The homogeneous equation of second degree is:
ax^{2} + 2hxy + by^{2} = 0.....................(2)
Comparing (1) and (2)
a = 1
h =2
b = 1
Now,
tan\(\theta\) = \(\frac {2\sqrt {h^2 - ab}}{a + b}\)
or, tan\(\theta\) = \(\frac {2\sqrt {(\frac {-5}2)^2 - 3 × 2}}{3 + 2}\)
or, tan\(\theta\) = \(\frac {2\sqrt {\frac {25}4 - \frac 61}}5\)
or, tan\(\theta\) = \(\frac {2\sqrt {\frac {25 - 24}4}}5\)
or, tan\(\theta\) = \(\frac {2 × \frac 12}{\frac 51}\)
or, tan\(\theta\) = \(\frac 15\)
or, \(\theta\) = tan^{-1}(\(\frac 15\))
∴ \(\theta\) = 11.31° _{Ans}
Here,
The given equation is:
12x^{2} - 23xy + 5y^{2} = 0.....................(1)
The homogeneous equation of second degree is:
ax^{2} + 2hxy + by^{2} = 0.....................(2)
Comparing (1) and (2)
a =12
h = -\(\frac {23}2\)
b = 5
Now,
tan\(\theta\) = ±\(\frac {2\sqrt {h^2 - ab}}{a + b}\)
or,tan\(\theta\) = ± \(\frac {2\sqrt {(\frac {-23}2)^2 - 12 × 5}}{12 + 5}\)
or,tan\(\theta\) = ± \(\frac {2\sqrt {\frac {529}4 - 60}}{17}\)
or,tan\(\theta\) = ± \(\frac {2\sqrt {\frac {529 - 240}4}}{17}\)
or,tan\(\theta\) = ± \(\frac {2\sqrt {289}}2\)× \(\frac 1{17}\)
or,tan\(\theta\) = ± \(\frac {17}{17}\)
∴tan\(\theta\) = ± 1
Taking +ve sign,
tan\(\theta\) = 1
or, tan\(\theta\) = tan 45°
∴ \(\theta\) = 45°
Taking -ve sign,
tan\(\theta\) = -1
or, tan\(\theta\) = tan (180 - 35)
or, tan\(\theta\) = tan 135°
∴ \(\theta\) = 135°
Hence, the obtuse angle between the pair of equation is: 135°. _{Ans}
Here,
The homogenous equation of second degree is:
ax^{2} + 2hxy +by^{2} = 0................................(1)
The given equation is:
2x^{2} - 5xy + 2y^{2} = 0...................................(2)
Comparing (1) and (2)
a = 2
2h = -5 i.e. h = -\(\frac 52\)
b = 2
If \(\theta\) be the angle between pair of lines:
tan\(\theta\) =± \(\frac {2\sqrt {h^2 - ab}}{a + b}\)
or, tan\(\theta\) =± \(\frac {2\sqrt {(\frac {-5}2)^2 - 2 × 2}}{2 + 2}\)
or, tan\(\theta\) =± \(\frac {2\sqrt {\frac {25}4 - \frac 41}}4\)
or, tan\(\theta\) =± \(\frac {\sqrt {\frac {25 - 16}4}}2\)
or, tan\(\theta\) =± \(\sqrt {\frac 94}\)× \(\frac 12\)
or, tan\(\theta\) =± \(\frac 32\)× \(\frac 12\)
or, tan\(\theta\) =± \(\frac 34\)
∴ \(\theta\) = tan^{-1}(± \(\frac 34\))
Hence, the acute angle (\(\theta\)) =tan^{-1}(± \(\frac 34\)) _{Ans}
Here,
Given equation is:
6x^{2} + 5xy - 3x - 2y - 6y^{2} = 0
or, 6x^{2} + 9xy - 4xy - 6y^{2} - 3x - 2y = 0
or, 3x (2x + 3y) - 2y (2x + 3y) - 1 (2x + 3y) = 0
or, (2x + 3y) (3x - 2y - 1) = 0
Either: 2x + 3y = 0..........................(1)
Or: 3x - 2y - 1 = 0............................(2)
Slope of equation (1), m_{1} = -\(\frac 23\)
Slope of equation (2), m_{2} = \(\frac 32\)
Again,
m_{1}× m_{2} = \(\frac {-2}3\)× \(\frac 32\) = -1
The product of two slopes = -1
Hence, these equations are perpendicular each other. _{Hence, Proved}
2x^{2} + 7xy + 3y^{2 } = 0
2x-3 =0 , x-3y = 0
2x+y =0 , x+3y = 0
2x+3 =0 , x+2y = 0
2x-y =0 , x-3y = 0
2x^{2 }- 5xy + 3y^{2 }= 0.
x+y =0 , 3x-2y =0
x-y =0 , 2x-3y =0
x-y =0 , 2x-0y =3
x+y =0 , 2x+3y =0
2x^{2 }+ xy- 6y^{2} = 0 .
1x-2y = 0 , x+3y = 0.
2x-3y = 0 , x+2y = 0.
2x+3y = 0 , x-2y = 0.
2x+3y = 0 , x+3y = 0.
X^{2 }+ x - y - y^{2 }= 0
x + y = 0 , x + 3y = 0
x - y = 2 , x + 3y = 0
x + y = 0 , y + 2x = 0
x - y = 0 , x + 2y = 0
3x^{2} - 5xy - 2y^{2 }- x + 2y = 0
x = 2y , 1x+y = 3
x = 2y , 3x+y = 1
x = 2y , 3x+y = 1
x = 1y , 3x+y = 2
Find the single equation which represents the pair of straight lines x=2 and y = 3
xy - 3x - 2y = 6 + 0
xy - 3x - 2y + 6 = 0
xy - 3x - 3y - 6 = 0
xy - 3x + 2y - 6 = 0
Find the homogeneous equation of 2nd degree from the pair of lines x = 2y&2x = y.
2x^{2 }- 5xy + 2y^{2 } = 0.
2x^{2 }- 2xy - 2y^{2 } = 0
2x^{2 }- 4xy + 3y^{2 } = 0.
5x^{2 }- 2xy + 2y^{2 } = 0.
6x^{2 }- xy - y^{2 }= 0
40(^0) , 145(^0)
45(^0) , 135(^0)
90(^0) , 135(^0)
45(^0) , 125(^0)
x^{2 }^{} - 3xy - 4y^{2}^{} = 0
99.04(^0) , 100.96(^0)
09.54(^0) , 126.96(^0)
θ = tan^{-1 }(± (frac{3}{5}) ) , 110.86(^0)
θ = tan^{-1 }(± (frac{5}{3}) ) or 59.04(^0) , 120.96(^0)
X^{2 }- 2xy cosec α + y^{2 } = 0
45(^o) - α
360(^o) - α
180(^o) - α
90(^o) - α
x^{2 }^{}^{}^{}^{}^{}- 4xy + y^{2 }= 0
360(^o)
180(^o)
120(^o)
145(^o)
3x^{2 }^{}- 7xy + 2y^{2 } = 0
135(^o)
180(^o)
190(^o)
360(^o)
x^{2} - 4xy + y^{2}^{} = 0
360(^o)
60(^o)
180(^o)
90(^o)
(frac{1}{2}) x^{2} + 2xy + (frac{1}{2}) y^{2 }
60(^o)
360(^o)
120(^o)
390(^o)
If the pair of lines represented by an equation mx^{2 }- 5xy - 6y^{2} = 0 are perpendicular to each other find the value of m.
12
6
5
4
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Rinku Bhattarai , Eden Garden Boarding Jhapa
Find the coordinate of the foot of the perpendicular from the point (6 , 8 ) to the line passing through the points (1 ,1 )and ( 9, 3)
Feb 01, 2017
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bishal thapa
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Feb 01, 2017
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Nijal
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Jan 11, 2017
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how to find eqn when the line is parallel and passing through points
Dec 31, 2016
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