 Practice Test | Kullabs.com
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135 (^0)
140(^0)
180(^0)
145(^0)
• ### What will be  the formula of the angle between the lines y = m1x+C1 and y = m2x + c2.

θ = (tan-^1) (frac{m1-m2}{±1+m1-m2})
θ = (tan-^1) (frac{m1+m2}{±1+m1+m2})
θ = (tan-^1) (frac{m2-m1}{±1-m1-m2})
θ = (tan-^1) (frac{m2-m1}{±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})

3
6
2
4
• ### What will be the slope of the straight line perpedicular to 4x+3y = 12.

(frac{3}{4})
(frac{2}{3})
(frac{4}{3})
(frac{6}{12})
• ### 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{2}{17})
(frac{15}{3})
(frac{15}{2})

5y+7x+13 = 0
5y+7x+13 = 0
5x+31+y7 = 0
5x+7y+31 = 0

9x+ y= 4
9x+ 9= y
9x+ y= 9
4x+ y= 9

4x+3y = 81
18x+4y = 0
3x+4y = 18
4x+ 3y= 9
• ### Find the equation of straight lines passing through a point (-6 , 4) and perpendicular to the line 3x-4y + 9 = 0 .

3x+4y+12 = 12
4y+3x+0 = 12
4x+3y+12 = 0
0+3y+12 = 4x

7x-5y = 44
7x+5y = 44
5x-7y = 44
44x-5y = 7x

3y-13-0 = 9x
13y+9x+3 = 0
3y+13+9x = 0
9x+3y+13 = 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-1y = 1 , 8-2y = 0
1-2xy = 0 , 2y-8 = x
2x-y = 1 , x+2y=8
1-2xy = 0 , 2y-8 = x
• ### 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{1x})-y = (sqrt{1})
y=0 and (sqrt{-3x})-y = (sqrt{-3})
y=0 and (sqrt{3x})-y = (sqrt{3})
y=0 and (sqrt{4x})-y = (sqrt{4})

0, 0
1, -1
1, 1
-1, -1