Straight Lines - Online Test
Equation of the line which perpendicular to the given line is x - y + k = 0
Since this line passes through (2,-3)
2-(-3) + k = 0
This implies k= -5
Hence the equation og the line is x - y =5
On solving the lines x+y=0 and x-y=5, we get the point of intersection as x = 5/2 and y = -5/2
Hence (5/2,-5/2) is the coordinates of orthogonal projection.
On solving the equations 8x+4y=1 and 4x+8y=3, we get the point of intersection as (-1/2,5/12)
On solving the equations 8x+4y=5 and 4x+8y=7, we get the point of intersection as (1/4,3/4)
On solving the equations 8x+4y=1 and 4x+8y=7, weget the point of intersection as (-5/12,13/12)
On solving the equations 8x+4y=5and 4x+8y=3, we the point of intersection as (7/12,1/12)
Let the points A(1/12,5/12), B(7/12,1/12) C(1/4,3/4) and D(-5/12,13/12) be the vertices of the quadrilaeral
Since the slopes of the opposite sides are equal the quadrilateral is a parallelogram
Slope of the diagonal AC is = 1
Slope of the diagonal BD is = -1
Since the product of the slopes is -1, the diagonals are perpendicular to each other
Hence the parallelogram is a rhombus
Let (x,0) be the point.
Then (by distance formula,
(x-2)2 +32 = c2
x2 -4x +4 + 9 - c2 = 0
x2 - 4x +13 - c2 = 0
x will be real if the b2 - 4ac 0
i.e; 16 - 4(13 - c2) 0
i.e; c2 - 9 0
i.e; |c| 3
Hence there is no point.
represents
On expansion we get
2[3(8-y) - 4(5 - x) +1(5y - 8x)]
On simplifying the equation 2( - 4x+2y+4) = 0 represents a striaght line.
The equations of the lines through ( 1 , 1 ) and making angles of with the line x + y = 0
If the lines make equal angles of 450 with the given line, x+y =0.
Then these lines must be perpendicular with each other.
This is possible only when the two lines are parallel to X axis and Y axis.
That is the equations should be x = a constant and y = a constant.
Since it passes through (1,1)
The equations should be x = 1 or x-1=0 and y=1 or y-1 =0
Thslope of the given line 2x+6y = 7 is -1/3
Hence the line which is parallel to the above line is
y = (-1/3)x+c
That is the y intercept is (0,c) and the x intercept is (3c,0)
Using the distance formula
d2 = (0-3c)2 + (3c-0)2
= 10c2
since the distance is 10 is given,
100 = 10c2
therefore c = 10
Since two values are possible, two lines can be drawn.
Expanding the given equation
px+2qx+py-3qy = p-q
px+py+2qx-3qy = p-q
p(x+y) -q(-2x+3y) = p-q
Equating the coeffiecients of like terms
x+y=1 and -2x+3y=1
On solving both the equations we get,
x = 2/5 and y = 3/5
Hence the line passes through the fixed point (2/5.3/5)
Equations of the lines are
x0, y0, then x+y=1
x0, y0, then x+y=-1
x0, y0, then x-y = 2
x0 , y0, then x-y=-2
Clearly these lines form a square, whose coordinates are (1,1),(1,-1),(-1,-1),(-1,1)
Hence its area is 4 x [1/2]1 x 1= 2 sq units
Since a,b,c are in A.P,
a+c = 2b
This implies a-2b+c = 0
This implies the the family of lines is concurrent at (1,2)
