On solving line 1 and line 2 we get x = -3 andy =4. Hence the point of intersection is (-3,4)
On solving line 2 and line 3 we get x = (-3/7) and y = 16/7. Hence the point of intersection is (-3/7, 16/7)
On solving line 3 and line 1 we get x = -3/5 and y - 8/5.Hence the point of intersection is (-3/5,8/5)
All the above points lie in the second quadrant. Hence the triangle formed by these lines also lie in the second quadrant.
Area of the triangle formed by the coordiates( x1,y1), (x2,y2) and (x3,y3) is given by
[x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)]
On solving the lines 1 and 2, the point of intersection is (0,0)
On solving the lines 2 and 3, the point of intersection is (4,8)
\On solving the line 3 and 1 , the point of intersection is (-2,-2)
Now substituting the values to find the area of the triangle,
Area = | [ 0 + 4(-2-0) +(-2)(0 - 8)] |
= 4 sq units
Let us take the coordinates as (-4,3), (2,-3) and (0,p).
If the points are collienear the |x1(y2 - y3) + x2(y3 - y1) + x3(y1-y2)| = 0
Now substituting the values |-4(-3 - p) +2(p -3) +0(3+3)|= 0
12 +4p +2p -6 +0 = 0
6p +6 = 0
6p = -6
Therefore p = -1
The required condition for concurrency is a3(b1c2 - b2c1) + b3(c1a2 - c2a1) + c3(a1b2 - a2b1) = 0
Here a1 = l, a2 =m, a3 = n and b1 = m, b2 = n, b3 = l and c1 = n, c2 = l and c3 = m
Substituting the values we get
n(ml - n2) + l(nm - l2) +m(ln - m2) = 0
This implies l3 + m3 + n3 - 3lmn = 0
That is (l + m+ n)(l2 + m2 + n2 -lm -mn - nl) = 0
This implies l + m + n = 0
The lines x+1=0 and y+1=0 are perpendicular to each other.
The slope of the line x+y =0 is -1
Hence the angle made by this line with respect to X axis is 450
In other words the angle made by this line with x+1=0 is 450
Clearly the other line with which it can make 450 is y+1=0
The lines are concurrent
On solving the lies 1 and 2 we get the point of intersection as (-1,2)
Similarly on solving lines 2 and 3, the point of intersection is (-1,2)
Similarly solving the lines 3 and 4, the point of intersection is (-1,2)
on solving lines 1 and 4 the point of intersection is (-1,2)
Since the point of intersection is the same for all the lines, the lines are concurrent.
The equation of the pair of straight lines paasing through the origin is fiven by is given by ax2 + 2hxy + by2 =0
The condition for perpendicularity between the pair of lines is a+b = 0
Hence the locus of the pair of these perpendicular straight lines should be xy = 0
The angle between two straight lines is given by
Here m1 = 2 and m2 = -2
Sustituting the values we get,
=
= 4/5 <600
when a point M which is in the 2nd quadrant is reflected in the origin, its image is formed in the 4th quadrant whose coordinates are (x,-y)
Hence the image of the point (1,-2) is (1,-2)