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### Straight 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 = $ax_1 + by_1 + c \over \sqrt{a^2+b^2}$

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

17. The lines represented by 5x2 + 3xy + ky2 = 0 are perpendicular to each other for k = [2074]

5

3

0

-5

→slope of second line is -a/b=5/2

12. The distance between the parallel lines y - 2x = 4 and 6x - y = 5 is[2070]

$1 \over{\sqrt{45}}$

$7 \over{\sqrt{45}}$

$17 \over{\sqrt{45}}$

$17 \over2{\sqrt{45}}$

I there is