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If PM is the perpendicular from P(2, 3) onto the line x + y = 3, then the coordinates of M are

1) (2, 1) 2) (-1, 4) 3) (1, 2) 4) (4, - 1) Solution: Let M(h, k) be the point on the line x + y = 3 So h+k = 3 …(i) Slope of line x + y =... View Article

The equation of the straight line perpendicular to 5x – 2y = 7 and passing through the point of intersection of the lines 2x + 3y = 1 and 3x + 4y = 6, is

1) 2x + 5y + 17 = 0 2) 2x + 5y - 17 = 0 3) 2x - 5y + 17 = 0 4) 2x - 5y = 17 Solution: Given lines 2x + 3y = 1 ..(i) 3x + 4y = 6 …(ii) By... View Article

If the point (a, a) falls between the line |x + y|= 4, then

1) |a|= 2 2) |a|= 3 3) |a| < 2 4) |a| < 3 Solution: Given |x + y| = 4 => x + y = 4 and x + y = -4 Since (a,a) lies between these... View Article

The equation of a line passing through (-2, -4) and perpendicular to line 3x – y + 5 = 0 is

1) 3y + x - 8 = 0 2) 3x + y + 6 = 0 3) x + 3y + 14 = 0 4) None of the above Solution: Given line is 3x - y + 5 = 0 => y = 3x + 5 Here... View Article

If the equations kx² – 2xy – y² – 2x + 2y = 0 represents a pair of lines, then k is equal to

1) 2 2) - 2 3) - 5 4) 3 Solution: Given kx2 - 2xy - y2 - 2x + 2y = 0 Comparing given equation with the general equation ax2 + 2hxy + by2 +... View Article

The straight line whose sum of the intercepts on the axes is equal to half of the product of the intercepts passes through the point

1) (1, 1) 2) (2, 2) 3) (3, 3) 4) (4, 4) Solution: Equation of the line in intercept form is x/a +y/b = 1 => bx + ay = ab …(i) Where a is... View Article

The centroid of the triangle formed by the pair of straight lines 12x² – 20xy + 7y² = 0 and the line 2x – 3y + 4 = 0 is

(1) (-7/3, 7/3) (2) (8/3, 8/3) (3) (-8/3, 8/3) (4) (4/3, 4/3) Solution: Given pair of straight lines 12x2 - 20xy + 7y2 = 0 => 12x2 - 6xy... View Article

If ax² – y² + 4x – y = 0 represents a pair of lines, then a is equal to

1) - 16 2) 16 3) 4 4) - 4 Solution: Given ax2 - y2 + 4x - y = 0 Comparing given equation with the general equation ax2 + 2hxy + by2 + 2gx +... View Article

The value of p for which the equation x² + pxy + y² – 5x – 7y + 6 = 0 represents a pair of straight lines, is

1) 5/2 2) 5 3) 2 4) 2/5 Solution: Given x2 + pxy + y2 - 5x - 7y + 6 = 0 Comparing given equation with the general equation ax2 + 2hxy + by2... View Article

The equation of pair of lines joining origin to the points of intersection of x² + y² = 9 and x + y = 3 is

1) (x2 + - x)2 = 9 2) xy = 0 3) (3 + y)2 + y2 = 9 4) (x - y)2 = 9 Solution: Given x2 + y2 = 9 ..(i) x + y = 3 …(ii) y = 3 - x Put y in (i)... View Article

The values of λ, for which the equation x² – y² – x + λy – 2 = 0 represents a pair of straight lines, is

1) -3, 1 2) -1, 1 3) 3, - 3 4) 3, 1 Solution: Given x2 - y2 - x + λy - 2 = 0 Comparing given equation with the general equation ax2 + 2hxy +... View Article

The three straight lines ax + by = c, bx + cy = a and cx + ay = b are collinear, if

1) b + c = a 2) c + a = b 3) a + b + c = 0 4) a + b = c Solution: Given ax + by = c ..(i) bx + cy = a ..(ii) cx + ay = b ..(iii) Adding... View Article

The equation of the line passing through the point of intersection of the lines x/a + y/b = 1 and x/b + y/a = 1 and having slope 0 is

1) x = a 2) y = b 3) y = ab/(a + b) 4) none of these Solution: Given x/a + y/b = 1 => bx + ay = ab ..(i) x/b + y/a = 1 => ax + by = ab... View Article

If (sin θ, cos θ) and (3, 2) lie on the same side of the line x + y = 1, then θ lies between

1) (0, π/2) 2) (0, π) 3) (π/4, π/2) 4) (0, π/4) Solution: Let L be given line x + y - 1 = 0 L(3, 2) => 3 + 2 -1 = 4 >0 L(sin θ, cos θ)... View Article

The point of concurrence of the line ax + by + c = 0 and a, b, c satisfy the relation 3a + 2b + 4c = 0 is

1) (3/2, 1/4) 2) (3/4, 1/4) 3) (3/4, 1/2) 4) (3/2, 1/2) Solution: Given relation 3a + 2b + 4c = 0 =>( ¾)a + (1/2) b + c = 0 …(i) Equation... View Article

If non – zero numbers a, b, c are in HP, then the straight line x/a + y/b + 1/c = 0 always passes through a fixed point. That point is

1) (1, - 1/2) 2) (1, -2) 3) (-1, -2) 4) (-1, 2) Solution: Given a, b, c are in HP. So 1/a, 1/b, 1/c are in AP. So 2/b = 1/a + 1/c 1/c = 2/b... View Article

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