\( \sqrt{34} \)
The distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by the distance formula: \[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] For the points \( (3, 4) \) and \( (7, 1) \): \[ x_1 = 3, \ y_1 = 4, \ x_2 = 7, \ y_2 = 1 \] \[ \text{Distance} = \sqrt{(7 - 3)^2 + (1 - 4)^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \] Thus, the distance is: \[ {5} \]
| List-I | List-II | ||
|---|---|---|---|
| (A) | $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ | (I) | $[\frac{1}{3} , 1 ]$ |
| (B) | $(x)=|[x]|,x \in [R$ | (II) | Z |
| (C) | $h(x) = |x - [x]| , x \in [R$ | (III) | W |
| (D) | $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ | (IV) | [0, 1) |
| (V) | { -1, 1} | ||
| List I | List II | ||
|---|---|---|---|
| (A) | $\lambda=8, \mu \neq 15$ | 1. | Infinitely many solutions |
| (B) | $\lambda \neq 8, \mu \in R$ | 2. | No solution |
| (C) | $\lambda=8, \mu=15$ | 3. | Unique solution |
A line \( L \) intersects the lines \( 3x - 2y - 1 = 0 \) and \( x + 2y + 1 = 0 \) at the points \( A \) and \( B \). If the point \( (1,2) \) bisects the line segment \( AB \) and \( \frac{a}{b} x + \frac{b}{a} y = 1 \) is the equation of the line \( L \), then \( a + 2b + 1 = ? \)
A line \( L \) passing through the point \( (2,0) \) makes an angle \( 60^\circ \) with the line \( 2x - y + 3 = 0 \). If \( L \) makes an acute angle with the positive X-axis in the anticlockwise direction, then the Y-intercept of the line \( L \) is?
If the slope of one line of the pair of lines \( 2x^2 + hxy + 6y^2 = 0 \) is thrice the slope of the other line, then \( h \) = ?