29
6
14
8
The locus of the point \( P(x, y) \), whose distance from the lines \( x + 2y + 7 = 0 \) and \( 2x - y + 8 = 0 \) is equal, is given by the equation:
\[ \frac{x + 2y + 7}{\sqrt{5}} = \pm \frac{2x - y + 8}{\sqrt{5}}. \]
Simplifying, we get:
\[ (x + 2y + 7)^2 = (2x - y + 8)^2. \]
For the combined equation of lines, we have:
\[ (x - 3y + 1)(3x + y + 15) = 0. \]
Expanding, we get:
\[ 3x^2 - 3y^2 - 8xy + 18x - 44y + 15 = 0. \]
Rewriting in standard form:
\[ x^2 - y^2 - \frac{8}{3}xy + 6x - \frac{44}{3}y + 5 = 0. \]
Thus, the equation becomes:
\[ x^2 - y^2 + 2hxy + 2gx + 2fy + c = 0, \]
where we identify:
\[ h = \frac{4}{3}, \quad g = 3, \quad f = -\frac{22}{3}, \quad c = 5. \]
Now, calculate \( g + c + h - f \):
\[ g + c + h - f = 3 + 5 + \frac{4}{3} + \frac{22}{3} = 8 + 6 = 14. \]
To solve the given problem, we need to understand the condition of equidistance from two lines given as:
The locus of points equidistant from two lines is the equation of the angle bisector, which can be derived from the given general second-degree equation:
\(E: x^2 - y^2 + 2hxy + 2gx + 2fy + c = 0\)
We rewrite the given line equations in a standard format:
The bisector of these two lines can be written using the formula:
\(\frac{|x + 2y - 8|}{\sqrt{1^2 + 2^2}} = \frac{|2x + y + 7|}{\sqrt{2^2 + 1^2}}\)
Squaring both sides, we get:
\(\frac{(x + 2y - 8)^2}{5} = \frac{(2x + y + 7)^2}{5}\)
This simplifies (after plugging into a homogeneous equation form for the second-degree equation) to:
\(x^2 - y^2 + 2 \cdot 0 \cdot xy + 2gx + 2fy + c = 0\) (for simplification in variable terms)
Since the expression is simple due to equidistance, by comparing the terms, \(h = 0\) and:
We are asked to find \(g + c + h - f\). Computing this gives:
\(g + c + 0 - (-14) = 14\)
Therefore, the value is 14.
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
Method used for separation of mixture of products (B and C) obtained in the following reaction is: 
The length of the perpendicular drawn from the point to the line is the distance of a point from a line. The shortest difference between a point and a line is the distance between them. To move a point on the line it measures the minimum distance or length required.
The following steps can be used to calculate the distance between two points using the given coordinates:
Note: If the two points are in a 3D plane, we can use the 3D distance formula, d = √(m2 - m1)2 + (n2 - n1)2 + (o2 - o1)2.
Read More: Distance Formula