Let the equation of two lines AB and CD be y = m_{1}x + c_{1 }and y = m_{2}x + c_{2} respectively.
let the lines AB and CD make angles θ_{1} and θ_{2} respectively with the positive direction of X-axis.
Then, tanθ_{1} = m_{1} and tanθ_{2} = m_{2}.
Let the lines AB and CD intersect each other at the point E.
Let the angles between the lines AB and CD
^{ ∠CEA = Φ}
Then by plane geometry, θ_{1} = Φ + θ_{2}
or,Φ =θ_{1}-θ_{2}
∴ tanΦ = tan(θ_{1 }-θ_{2}) = \(\frac {tan\theta_1 - tan\theta_2}{1 + tan\theta_1 tan\theta_2}\) = \(\frac{m_1 - m_2}{1+tan\theta_1 tan\theta_2}\) .........(i)
Again, let ∠BAC = Ψ
Then by place geometry ,Φ + Ψ = 180^{0}
or, Ψ = 180^{0} - Φ
or, tan Ψ = tan (180 - Φ) = -tan Φ = - \(\frac{m_1 - m_2}{1+m_1m_2}\) ............(ii)
Hence if angles between the lines y = m_{1}x + c_{1} and y = m_{2}x + c_{2} be the θ then,
tanθ =± \(\frac{m_1 - m_2}{1+m_1m_2}\)
θ = tan^{-1}(± \(\frac{m_1 - m_2}{1+m_1m_2}\))
Condition of Perpendicularity
Two lines AB and CD will be perpendicular to each other if the angle between them θ = 90^{o}.
We have tanθ = ± \(\frac{m_1 - m_2}{1+m_1m_2}\)
or, tan90^{0} = ± \(\frac{m_1 - m_2}{1+m_1m_2}\)
or, cot90^{0} = ± \(\frac{1+m_1m_2}{m_1 - m_2}\)
or, 0 = \(\frac{1+m_1m_2}{m_1 - m_2}\)
or, 1 +m_{1}m_{2} =0
or, m_{1}m_{2} = -1
Two lines will be perpendicular to each other if m_{1}m_{2} = -1
i.e. if product of the slopes = -1 .
Condition of Parallelism
Two lines AB and CD will be parallel to each other if the angle between them θ = 0^{0}.
we have, tanθ = ± \(\frac{m_1 - m_2}{1+m_1m_2}\)
or, tan0^{0} = ± \(\frac{m_1 - m_2}{1+m_1m_2}\)
or, 0 = \(\frac{m_1 - m_2}{1+m_1m_2}\)
or, m_{1} - m_{2} = 0
or, m_{1 }= m_{2}
∴ Two lines will be parallel to each other if m_{1} = m_{2} i.e. if slopes are equal.
Let equations of two straight lines AB and CD be A_{1}x + B_{1y + }C_{1} = 0 and A_{2}x + B_{2}y + C_{2} = 0 respectively.
Then slope of AB = -\(\frac{A_1}{B_1}\)
Slope of CD = -\(\frac{A_2}{B_2}\)
Let the lines AB and CD make angles θ_{1} and θ_{2} with the positive direction of X-axis.
Then, tanθ_{1} = -\(\frac{A_1}{B_1}\) and tanθ_{2} = -\(\frac{A_2}{B_2}\)
Let ∠CEA = Φ.
Then, θ_{1} = θ_{1} - θ_{2}
or, Φ = θ_{1} - θ_{2}
∴ tan Φ = tan( θ_{1} - θ_{2}) = \(\frac{tan\theta_1 - tan\theta_2}{1 + tan\theta_1 tan\theta_2}\) = \(\frac{A_2 B_1 - A_1 B_2}{A_1 A_2 + B_1 B_2}\) = -\(\frac{A_1 B_2 - A_2 B_1}{A_1 A_2 + B_1 B_2}\) ......(i)
Let ∠BEC = Ψ
Then Ψ + Φ = 180^{0}
or, Ψ = 180^{0} - Φ
∴ tan Ψ = tan(180^{0} - Φ) = -tanΦ = \(\frac{A_1 B_2 - A_2 B_1}{A_1 A_2 + B_1 B_2}\) .........(ii)
Hence if angles between the lines A_{1}x + B_{1y + }C_{1} = 0 and A_{2}x + B_{2}y + C_{2} = 0 is θ, then
tanθ = ± \(\frac{A_1 B_2 - A_2 B_1}{A_1 A_2 + B_1 B_2}\)
or, θ = tan^{-1 }(±\(\frac{A_1 B_2 - A_2 B_1}{A_1 A_2 + B_1 B_2}\) )
Condition of Perpendicularity
Two lines AB and CD will be perpendicular to each other if θ = 90^{0}
Then tan90^{0} = ± \(\frac{A_1 B_2 - A_2 B_1}{A_1 A_2 + B_1 B_2}\)
or, ∞ = \(\frac{A_1 B_2 - A_2 B_1}{A_1 A_2 + B_1 B_2}\)
∴ A_{1}A_{2} + B_{1}B_{2} = 0
Condition of Parallelism
Two lines AB and CD will be parrallel to each other if θ = 0^{0}
Then, tan0^{0} = ± \(\frac{A_1 B_2 - A_2 B_1}{A_1 A_2 + B_1 B_2}\)
or, 0 = \(\frac{A_1 B_2 - A_2 B_1}{A_1 A_2 + B_1 B_2}\)
or, A_{1}B_{2} - A_{2}B_{1} = 0
or, A_{1}B_{2 }= A_{2}B_{1}
∴ \(\frac{A_1}{A_2}\) = \(\frac{B_1}{B_2}\)
Equation of the given line is ax + by + c = 0
Slope of this line = -\(\frac{coefficient \;of \;x}{coefficient \;of \;y}\) = -\(\frac{a}{b}\)
Slope of the line parallel to this line = -\(\frac{a}{b}\)
Now, equation of a line having slope -\(\frac{a}{b}\) is given by
y = mx + c
or, y = -\(\frac{a}{b}\)x + c
or, by = -ax + bc
or, ax + by - bc = 0
or, ax + by + k = 0 where, k = -bc.
Hence equation of any line parallel to ax + by + c = 0 is given by ax + by + k = 0 where k is an arbitrary constant.
Equation of the given line is ax + by + c = 0
Slope of this line = -\(\frac{coefficient\;of\;x}{coefficient\;of\;y}\) = -\(\frac{a}{b}\)
Slope of the line perpendicular to given line = \(\frac{b}{a}\)
Now equation of a line having slope \(\frac{b}{a}\) is given by
y = mx + c
or, y = \(\frac{b}{a}\)x + c
or, ay = bx + ac
or, bx - ay + ac = 0
or, bx - ay +k = 0 where k = ac.
Hence, equation of any line perpendicular to ax + by + c = 0 is given by bx - ay + k = 0 where k is an arbitrary constant.
Angle between the lines y = m_{1 }+ c_{1} and y = m_{2 + }c_{2} and y = m_{2}x + c_{2}
Condition of perpendicularity
m_{1}m_{2} = -1
Condition of Parallelism
m_{1 }= m_{2}
Angle between the lines A_{1}x + B_{1}y + C_{1} = 0 and A_{2}x + B_{2}y +C_{2} = 0
Condition of perpendicularity
A_{1}A_{2} + B_{1}B_{2} = 0
Condition of Parallelism
A_{1}B_{2 }= A_{2}B_{1}
Equation of any line parallel to ax + by + c = 0
k = -bc
Equation of any line perpendicular to ax +by +c = 0
k = ac
.
Given:
a_{1}x + b_{1}y + c_{1} = 0..............................(1)
a_{2}x + b_{2}y + c_{2} = 0..............................(2)
Slope of eq^{n} (1), m_{1} = - \(\frac {x-coefficient}{y-coefficient}\) = -\(\frac {a_1}{b_1}\)
Slope of eq^{n} (2), m_{2} = - \(\frac {x-coefficient}{y-coefficient}\) = -\(\frac {a_2}{b_2}\)
when the lines are parallel, then:
m_{1} = m_{2}
or,-\(\frac {a_1}{b_1}\) =-\(\frac {a_2}{b_2}\)
∴ a_{1}b_{2} = a_{2}b_{1 Ans}
when the lines are perpendicular, then:
m_{1}× m_{2}= - 1
or,-\(\frac {a_1}{b_1}\)×-\(\frac {a_2}{b_2}\) = - 1
∴ a_{1}a_{2} = - b_{1}b_{2}_{Ans}
Prove that the two straight lines 5x + 4y - 10 = 0 and 15x + 12y - 7 = 0 are parallel to each other.
Here,
5x + 4y - 10 = 0............................(1)
15x + 12y - 7 = 0.........................(2)
Slope of eq^{n} (1), m_{1} = - \(\frac {x-coefficient}{y-coefficient}\) = -\(\frac 54\)
Slope of eq^{n} (2), m_{2} = - \(\frac {x-coefficient}{y-coefficient}\) = -\(\frac {15}{12}\) = -\(\frac 54\)
m_{1} = m_{2} = - \(\frac 54\)
∴ The given two lines are parallel to each other. _{Proved}
Here,
x + 3y = 2..................................(1)
6x - 2y = 9................................(2)
Slope of eq^{n} (1), m_{1} = - \(\frac {x-coefficient}{y-coefficient}\) = - \(\frac 13\)
Slope of eq^{n} (2), m_{2} = - \(\frac {x-coefficient}{y-coefficient}\) = - \(\frac 6{-2}\) =3
We have,
m_{1}× m_{2} =- \(\frac 13\)× 3 = - 1
∴m_{1}× m_{2} = - 1
Hence, the given two lines are perpendicular each other. _{Proved}
Here,
3x - 2y - 5 = 0..............................(1)
2x + py - 3 = 0.............................(2)
Slope of eq^{n} (1), m_{1} = - \(\frac {x-coefficient}{y-coefficient}\) = - \(\frac 3{-2}\) = \(\frac 32\)
Slope of eq^{n} (2), m_{2} = - \(\frac {x-coefficient}{y-coefficient}\) = - \(\frac 2{p}\)
When two lines are parallel lines, then:
m_{1} = m_{2}
or,\(\frac 32\) = - \(\frac 2{p}\)
or, p = \(\frac {-2 × 2}3\)
∴ p = -\(\frac 43\) _{Ans}
Here,
4x + ky - 4 = 0..............................(1)
2x - 6y = 5.............................(2)
Slope of eq^{n} (1), m_{1} = - \(\frac {x-coefficient}{y-coefficient}\) = - \(\frac 4k\)
Slope of eq^{n} (2), m_{2} = - \(\frac {x-coefficient}{y-coefficient}\) = - \(\frac 2{-6}\) = \(\frac 13\)
when two line are perpendicular to each other,
m_{1}× m_{2} = - 1
or, - \(\frac 4k\) × \(\frac 13\) = - 1
or, - \(\frac 43\)× - 1 = k
∴ k = \(\frac 43\) _{Ans}
The formulae of angle between y = m_{1}x + c_{1} andy = m_{2}x + c_{2}is:
tan\(\theta\) = ± \(\frac {m_1 - m_2}{1 + m_1m_2}\)
when m_{1}× m_{2} = -1, the two lines are perpendicular to each other.
when m_{1} = m_{2}, the two lines are parallel to each other.
Here,
Slope of points (3, -4) and (-2, a)
m_{1} = \(\frac {y_2 - y_1}{x_2 - x_1}\) = \(\frac {a + 4}{-2 - 3}\) = \(\frac {-(a + 4)}5\)
Given eq^{n} is y + 2x + 3 = 0
Slope of above eq^{n} (m_{2}) = - \(\frac {x-coefficient}{y-coefficient}\) = -\(\frac 21\) = - 2
when lines are parallel then,
m_{1} = m_{2}
or,\(\frac {-(a + 4)}5\) = - 2
or, a + 4 = 10
or, a = 10 - 4
∴ a = 6 _{Ans}
Here,
Slope of the points (3, -4) and (-2, 6) is:
m_{1} = \(\frac {y_2 - y_1}{x_2 - x_1}\) = \(\frac {6 + 4}{-2 - 3}\) = \(\frac {10}{-5}\) = -2
Slope of the eq^{n} y + 2x + 3 = 0 is:
m_{2} = - \(\frac {x-coefficient}{y-coefficient}\) = - \(\frac 21\) = -2
From above,
m_{1}= m_{2} = - 2
Hence, the lines are parallel. _{Proved}
Here,
Given eq^{n} is kx - 3y + 6 = 0
Slope of above eq^{n} is:
m_{1} = - \(\frac {x-coefficient}{y-coefficient}\) = - \(\frac k{-3}\) = \(\frac k3\)
Slope ofthe point (4, 3) and (5, -3) is:
m_{2} = \(\frac {y_2 - y_1}{x_2 - x_1}\) = \(\frac {-3 - 3}{5 - 4}\) = - \(\frac 61\) = -6
If lines are perpendicular then:
m_{1}× m_{2} = -1
or, \(\frac k3\)× -6 = -1
or, k = \(\frac {-1}{-2}\)
∴ k = \(\frac 12\) _{Ans}
Here,
Given equations of the lines are:
y = m_{1}x + c_{1}...............................(1)
y = m_{2}x + c_{2}...............................(2)
If \(\theta\) be the angle between two lines (1) and (2);
The formula of angle between the given lines is:
tan\(\theta\) =± \(\frac {m_1 - m_2}{1 + m_1m_2}\)
∴ \(\theta\) = tan^{-1}(± \(\frac {m_1 - m_2}{1 + m_1m_2}\))
If two lines are perpendicular (\(\theta\) = 90°)
tan 90° =± \(\frac {m_1 - m_2}{1 + m_1m_2}\)
or,∞=± \(\frac {m_1 - m_2}{1 + m_1m_2}\)
or, \(\frac 10\) =± \(\frac {m_1 - m_2}{1 + m_1m_2}\)
or, 1 + m_{1}m_{2} = 0
∴ m_{1}m_{2} = -1 _{Ans}
Here,
The given equations are:
2x + ay + 3 = 0....................(1)
3x - 2y = 5.............................(2)
Slope of equation (1), m_{1} = -\(\frac {x-coefficient}{y-coefficient}\) = - \(\frac 2a\)
Slope of equation (2), m_{2} = -\(\frac {x-coefficient}{y-coefficient}\) = - \(\frac 3{-2}\) = \(\frac 32\)
If equation (1) and equation (2) are perpendicular to each other:
m_{1}× m_{2} = -1
or,- \(\frac 2a\)×\(\frac 32\) = -1
or, -6 = - 2a
or, a = \(\frac 62\)
∴ a = 3 _{Ans}
Here,
Given equations of the lines are:
2x + 4y - 7 = 0...........................(1)
6x + 12y + 4 = 0.......................(2)
Slope of equation (1) is: m_{1} = - \(\frac {x-coefficient}{y-coefficient}\) = -\(\frac 24\) = -\(\frac 12\)
Slope of equation (2) is: m_{2} = - \(\frac {x-coefficient}{y-coefficient}\) = -\(\frac 6{12}\) = -\(\frac 12\)
∴ m_{1} = m_{2} = \(\frac {-1}2\)
Since, the slope of these equations are equal, the lines are parallel to each other. _{Proved}
Given lines are:
3x + 5y = 7 i.e. 3x + 5y - 7 = 0........................(1)
3y = 2x + 4 i.e. 2x - 3y + 4 = 0........................(2)
Slope of eq^{n} (1), m_{1} = - \(\frac {x-coefficient}{y-coefficient}\) = \(\frac {-2}{-3}\) = \(\frac 23\)
Slope of eq^{n} (2), m_{2} = - \(\frac {x-coefficient}{y-coefficient}\) = \(\frac {-3}{5}\)
Let \(\theta\) be the angle between the equation (1) and (2):
tan\(\theta\) =± \(\frac {m_1 - m_2}{1 + m_1m_2}\)
or, tan\(\theta\) =± (\(\frac {\frac 23 + \frac 35}{1 - \frac 23 × \frac 35}\))
or, tan\(\theta\) =± (\(\frac {\frac {10 + 9}{15}}{\frac {5 - 2}5}\))
or, tan\(\theta\) =± (\(\frac {19}{15}\) × \(\frac 53\))
or, tan\(\theta\) =± \(\frac {19}9\)
For acute angle,
tan\(\theta\) = \(\frac {19}9\)= 2.11
∴ \(\theta\) = 65°
∴ The acute angle between two lines is 65°. _{Ans}
Here,
Given equation are:
3y - x - 6 = 0..............................(1)
y = 2x + 5 i.e. -2x + y = 5...................(2)
Slope of eq^{n} (1), m_{1}= -\(\frac {x-coefficient}{y-coefficient}\) = -\(\frac {(-1)}3\) = \(\frac 13\)
Slope of eq^{n} (2), m_{2}= -\(\frac {x-coefficient}{y-coefficient}\) = -\(\frac {(-2)}1\) = 2
If \(\theta\) be the angle between the eq^{n} (1) and (2),
tan\(\theta\) =± \(\frac {m_1 - m_2}{1 + m_1m_2}\)
or, tan\(\theta\) =± \(\frac {\frac 13 - 2}{1 + \frac 13 × 2}\)
or, tan\(\theta\) =± \(\frac {\frac {1 - 6}3}{\frac {3 + 2}3}\)
or, tan\(\theta\) =± \(\frac {-5}3\)× \(\frac 35\)
∴ tan\(\theta\) =± (-1)
Taking -ve sign,
tan\(\theta\) = +1
tan\(\theta\) = 45°
∴\(\theta\) = 45° _{Ans}
Here,
x - 3y = 4......................(1)
2x - y = 3......................(2)
Slope of eq^{n} (1), m_{1} = -\(\frac {x-coefficient}{y-coefficient}\) = \(\frac {-1}{-3}\) = \(\frac 13\)
Slope of eq^{n} (2), m_{2} = -\(\frac {x-coefficient}{y-coefficient}\) = \(\frac {-2}{-1}\) = 2
If \(\theta\) be the angle between the eq^{n} (1) and (2),
tan\(\theta\) =± \(\frac {m_1 - m_2}{1 + m_1m_2}\)
or, tan\(\theta\) =± \(\frac {\frac 13 - 2}{1 + \frac 13 × 2}\)
or, tan\(\theta\) =± \(\frac {\frac {1 - 6}3}{\frac {3 + 2}3}\)
or, tan\(\theta\) =± \(\frac {-5}3\)× \(\frac 35\)
∴ tan\(\theta\) =± (-1)
Taking -ve sign,
tan\(\theta\) = +1
tan\(\theta\) = 45°
∴\(\theta\) = 45° _{Ans}
Here,
Given equation are:
y - 3x - 2 = 0
or, -3x + y - 2 = 0..............................(1)
y = 2x + 5
or, - 2x + y = 5...................................(2)
Slope of eq^{n} (1), m_{1} = -\(\frac {x-coefficient}{y-coefficient}\) = \(\frac {-3}{-1}\) = 3
Slope of eq^{n} (2), m_{2} = -\(\frac {x-coefficient}{y-coefficient}\) = -\(\frac {-2}{1}\) =2
If \(\theta\) be the angle between two lines,
tan\(\theta\) =± \(\frac {m_1 - m_2}{1 + m_1m_2}\)
or, tan\(\theta\) =± \(\frac {3 - 2}{1 + 3 × 2}\)
or, tan\(\theta\) =± \(\frac 1{1 + 6}\)
∴ tan\(\theta\) =± \(\frac 17\)
Taking +ve sign,
\(\theta\) = tan^{-1}(\(\frac 17\))
∴ \(\theta\) = 8.13° _{Ans}
Here,
Given lines are:
x = 3y + 8
i.e. x - 3y - 8 = 0..................................(1)
2x + 11 = 7y
i.e. 2x - 7y + 11 = 0............................(2)
Slope of eq^{n }(1),^{}m_{1}= - \(\frac {x-coefficient}{y-coefficient}\) = - \(\frac 1{-3}\) = \(\frac 13\)
Slope of eq^{n }(2),^{}m_{2}= - \(\frac {x-coefficient}{y-coefficient}\) = - \(\frac 2{-7}\) = \(\frac 27\)
Let \(\theta\) be the angle between the equation (1) and (2),
tan\(\theta\) =± \(\frac {m_1 - m_2}{1 + m_1m_2}\)
or, tan\(\theta\) = ± \(\frac {\frac 13 - \frac 27}{1 + \frac 13 × \frac 27}\)
or, tan\(\theta\) =± \(\frac {\frac {7 - 6}{21}}{\frac {21 + 2}{21}}\)
or, tan\(\theta\) =± \(\frac 1{21}\)× \(\frac {21}{23}\)
∴ tan\(\theta\) =± \(\frac 1{23}\)
For obtuse angle,
tan\(\theta\) = -\(\frac 1{23}\)
or, tan\(\theta\) = tan (180° -2°)
∴ \(\theta\) = 178°
∴ The obtuse angle between two lines is 178°. _{Ans}
What will be the obtuse angle between two lines 2x - y + 4 = 0 and 3x+y+3 = 0 .
140(^0)
180(^0)
145(^0)
135 (^0)
What will be the formula of the angle between the lines y = m_{1x}+C_{1 }and y = m_{2}x + c_{2}.
θ = (tan-^1) (frac{m2-m1}{±1-m1+m2})
θ = (tan-^1) (frac{m2-m1}{±1-m1-m2})
θ = (tan-^1) (frac{m1+m2}{±1+m1+m2})
θ = (tan-^1) (frac{m1-m2}{±1+m1-m2})
If the straight lines px+3y-12 = 0 and 4y-3x+7 = 0 are parallel to each other , find the value of P.
(frac{9}{4})
(frac{3x}{4y})
(frac{-9}{4})
(frac{-9}{-4})
If the line passing through (3 , -4) and (-2 , a) is parallel to the line given by the equation y+2x+3 = 0 , what will be the value of a.
4
6
2
3
What will be the slope of the straight line perpedicular to 4x+3y = 12.
(frac{2}{3})
(frac{4}{3})
(frac{6}{12})
(frac{3}{4})
If the straight lines 2x+3y+6 = 0 and ax-5y+20 = 0 are perpendicular to each other , find the value of a .
(frac{3}{4})
(frac{15}{2})
(frac{15}{3})
(frac{2}{17})
Find the equation of a straight line which is parallel to the line with the equation 5x+7y = 14 and passes through to the point (-2 , -3) .
5x+7y+31 = 0
5x+31+y7 = 0
5y+7x+13 = 0
5y+7x+13 = 0
Find the equation of a straight line which passes through the point (2,1) and is parallel to the line joining the points (2,3) and (3,-1) .
9x+ y= 9
9x+ y= 4
9x+ 9= y
4x+ y= 9
Find the equation of straight lines passing through the point (2 , 3) and perpendicular to the line 4x-3y = 10.
4x+ 3y= 9
3x+4y = 18
18x+4y = 0
4x+3y = 81
Find the equation of straight lines passing through a point (-6 , 4) and perpendicular to the line 3x-4y + 9 = 0 .
4y+3x+0 = 12
3x+4y+12 = 12
4x+3y+12 = 0
0+3y+12 = 4x
Find the equation of straight lines passing through a point (7 , 1) and perpendicular to the line 5x+7y+12 = 0.
5x-7y = 44
7x-5y = 44
44x-5y = 7x
7x+5y = 44
The point C divides the line segment AB joining the points A (2 , 3) and B(-4) in the ratio 2:1 . Find the equation of the line passing through the point C and perpendicular to AB .
3y-13-0 = 9x
3y+13+9x = 0
9x+3y+13 = 0
13y+9x+3 = 0
Find the equation of the straight lines passing through the point (2 , 3) and making an angle of 45(^0) with the line x-3y = 2.
2x-y = 1 , x+2y=8
1-2xy = 0 , 2y-8 = x
1-2xy = 0 , 2y-8 = x
2x-1y = 1 , 8-2y = 0
Find the equation of the straight lines passing through the point (1,0) & inclined at an angle of 30(^0) with the line x- (sqrt{3y}) = 4.
y=0 and (sqrt{4x})-y = (sqrt{4})
y=0 and (sqrt{1x})-y = (sqrt{1})
y=0 and (sqrt{3x})-y = (sqrt{3})
y=0 and (sqrt{-3x})-y = (sqrt{-3})
If the line (frac{x}{a}) + (frac{y}{b}) = 1 passes through the point of intersection of the lines x+y = 3 and 2x-3y = 1 and is parallel to the line y=x-6 , then find the values of a and b.
1, -1
0, 0
1, 1
-1, -1
You must login to reply
Apr 20, 2017
0 Replies
Successfully Posted ...
Please Wait...
Apr 20, 2017
0 Replies
Successfully Posted ...
Please Wait...
Find the equation of the straight line passing through the point (5,0)and making 45° with the line 4x-5y 9=0
Find the equation of the straight line passing through the point(5,0)and making 45°withthe line 4x-5y 9=0
Mar 06, 2017
0 Replies
Successfully Posted ...
Please Wait...
lines 2x-3ay=4 and x 2y=1 are perpendicular to each other.
Lines 2x-3ay=4 and x 2y=1 are perpendicular to each other.
Jan 26, 2017
1 Replies
Successfully Posted ...
Please Wait...