csitfamilyStraight Line
16. The perpendicular distance from (-1,-2) to the line 3x + 4y + 12 = 0 is [2075]
-1
1
2
3
Solutions
The distance of any point (x1,y1) from the line ax+by+c=0 is:
Distance = $ ax_1 + by_1 + c \over \sqrt{a^2+b^2}$
Distance = $ ax_1 + by_1 + c \over \sqrt{a^2+b^2}$
17. If the lines represented by ky2 + (2 - k)2xy - 3x2 = 0 are perpendicular if k = [2075]
-3
3
0
1
16. The line kx + 3y + 5 = 0 and 5x - 2y - 6 = 0 are perpendicular to each other for k = [2074]
6/5
-6/5
3/5
-3/5
Solutions
Given kx+3y+5=0 and 5x-2y-6=0 and are perpendicular to each other
→slope of first line is -a/b=-k/3
since they are perpendicular ,the product of slopes should be -1
-k/3 x 5/2 = -1
-5k/6 = -1
k = 6/5
→slope of first line is -a/b=-k/3
since they are perpendicular ,the product of slopes should be -1
-k/3 x 5/2 = -1
-5k/6 = -1
k = 6/5
17. The lines represented by 5x2 + 3xy + ky2 = 0 are perpendicular to each other for k = [2074]
5
3
0
-5
12. The distance between the parallel lines y - 2x = 4 and 6x - y = 5 is[2070]
$ 1 \over{\sqrt{45}}$
$ 7 \over{\sqrt{45}}$
$ 17 \over{\sqrt{45}}$
$ 17 \over2{\sqrt{45}}$
