The equation of a straight line; perpendicular to 8x - 4y = 6 and forming a triangle of area 6sq. units with coordinate axes, is

• x - 2y = 6

• 4x + 3y = 12

• 4x + 3y + 24 = 0

• 3x + 4y = 12

If the image of $\left(\frac{- 7}{6}, \frac{- 6}{5}\right) \mathrm{in} \mathrm{a} \mathrm{line} \mathrm{is} \left(1, 2\right), \mathrm{then} \mathrm{the} \mathrm{equation} \mathrm{of} \mathrm{the} \mathrm{line} \mathrm{is}$

• 4x + 3y = 1

• 3x - y = 0

• 4x - y = 0

• 3x + 4y = 1

If a line l passes through (k, 2k), (3k, 3k) and (3, 1), k $\ne$ 0, then the distance from the origin to the line is

• $\frac{1}{\sqrt{5}}$

• $\frac{4}{\sqrt{5}}$

• $\frac{3}{\sqrt{5}}$

• $\frac{2}{\sqrt{5}}$

The area (in sq. units) of the triangle formed by the lines x² - 3y + y = 0 and x + y + 1 = 0, is

• $\frac{2}{\sqrt{3}}$

• $\frac{\sqrt{3}}{2}$

• $5\sqrt{2}$

• $\frac{1}{2\sqrt{5}}$

If ${\mathrm{x}}^2 + {\mathrm{\alpha y}}^2 +\hspace{0.17em}2\mathrm{\beta y} = {\mathrm{\alpha }}^2$ represents a pair of perpendicular lines, then p equals to

• 4a

• a

• 2a

• 3a

A circle with centre at (2, 4) is such that the line x + y + 2 = 0 cuts a chord of length 6. The radius of the circle is

• $\sqrt{41} \mathrm{cm}$

• $\sqrt{11} \mathrm{cm}$

• $\sqrt{21} \mathrm{cm}$

• $\sqrt{31} \mathrm{cm}$

The point at which the circles x² + y² - 4x - 4y + 7 = 0 and x² + y² - 12x - 10y + 45 = 0 touch each other, is

• $\left(\frac{13}{5}, \frac{{\displaystyle 14}}{{\displaystyle 5}}\right)$

• $\left(\frac{2}{5}, \frac{{\displaystyle 5}}{{\displaystyle 6}}\right)$

• $\left(\frac{14}{5}, \frac{{\displaystyle 13}}{{\displaystyle 5}}\right)$

• $\left(\frac{12}{5}, 2 + \frac{{\displaystyle \sqrt{21}}}{{\displaystyle 5}}\right)$

The condition for the lines lx + my + n = 0 and l₁a + m₁y + n₁ = 0 to be conjugate with respect to the circle x² + y² = r², is

• r²(ll₁ + mm₁) = nn₁

• r²(ll₁ - mm₁) = nn₁

• r²(ll₁ + mm₁) + nn₁ = 0

• r²(lm₁ + l₁m) = nn₁

The length of the common chord of the two circles x² + y² - 4y = 0 and x² + y² - 8x - 4y + 11 = 0, is

• $\frac{\sqrt{145}}{4} \mathrm{cm}$

• $\frac{\sqrt{11}}{2} \mathrm{cm}$

• $\sqrt{135} \mathrm{cm}$

• $\frac{\sqrt{135}}{4}$

The locus of the centre of the circle, which cuts the circle x² + y² - 20 + 4 = 0 orthogonally and touches the line x = 2, is

• x² = 16y

• y² = 4x

• y² = 16x

• x² = 4y