Question

The equation of straight line equally inclined to the axes and equidistant from the points \[(1,-2)\]and \[(3,4)\] is \[ax+by+c=0\] where
\[1)a=1,b=1,c=1\]
\[2)a=1,b=-1,c=-1\]
\[3)a=1,b=1,c=2\]
\[4)\text{ None of these}\]

Answer 1

Hint : In this question firstly find out the \[slope(m)\] of the given equation then find out the mid points of the equation. After that find out the equation of the line passing through the given points then compare the obtained equation with general equation of straight line to obtain the given result.

Complete step-by-step solution:
A line is the locus of the two points such that every point of the path joining two points lies on that locus.
A straight line or line is an endless \[\text{one-dimensional}\] figure that has no width and it does not have any curve, it can be horizontal, vertical or slanted.
General Equation of a Straight Line:
The general equation of a straight line is given as \[ax+by+c=0\] where,
\[\text{a,b,c}\] are constants , \[\text{x,y}\] are variables
\[\text{Slope}\] intercept form of an equation of line:
A straight line has \[\text{slope}\] \[\text{(m=tan}\theta \text{)}\], where \[\theta \] is the angle formed by the line with positive \[\text{x-axis}\], and \[\text{y-intercept}\] as \[\text{c}\] which can be written as \[\text{y=mx+c}\].
Three points \[A,B,C\] are collinear if and only if \[\text{slope of AB= slope of BC}\]
If the straight line is formed by positive \[\text{x-axis}\]and have the \[\text{Slope}\]\[\text{(m=tan}\theta \text{)}\]and the line is passing through the point \[\text{(}{{\text{x}}_{1}}\text{,}{{\text{y}}_{1}})\] then the equation will be :
\[y-{{y}_{1}}=m(x-{{x}_{1}})\]
To calculate the mid point find the exact center point between two defined points in aline segment. We may use the following formula to calculate the mid point that bisects the line segment
\[\text{Midpoint=( }\dfrac{{{\text{x}}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2})\]
Now according to the question:
The equation of straight line equally inclined to the axes means that the \[\text{Slope}\] of the line will be
\[\text{m=tan}\theta \]
\[\text{m=tan4}{{\text{5}}^{\circ }}\]
\[\text{m=1}\]
Now the lines are passing through mid point \[(1,-2)\]and \[(3,4)\]
Hence mid points are \[(\dfrac{{{\text{x}}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2})\]
\[{{\text{x}}_{1}}=1,{{y}_{1}}=-2\] and \[{{x}_{2}}=3,{{y}_{2}}=4\]
\[\Rightarrow (\dfrac{1+3}{2},\dfrac{-2+4}{2})\]
\[=(\dfrac{4}{2},\dfrac{2}{2})\]
\[=(2,1)\]
Mid points\[=(2,1)\]
Hence, the equation of line passes through \[(2,1)\] and \[\text{Slope (m=1)}\]
The equation of line will be given by : \[y-{{y}_{1}}=m(x-{{x}_{1}})\]
Where \[{{\text{x}}_{1}}=2\] , \[{{y}_{1}}=1\] and \[\text{m=1}\]
\[\Rightarrow y-1=1(x-2)\]
\[\Rightarrow y-1=x-2\]
\[\Rightarrow y-x+1\]
\[\Rightarrow x-y-1=0\]
Comparing the equation \[x-y-1=0\] by the general equation of straight line \[ax+by+c=0\]
We get the value of \[\text{a,b,c}\]
So, \[a=1,b=-1,c=-1\]
From the above process we can conclude that option \[(2)\] is correct.

Note:We must keep one thing in mind that in the calculation of \[\text{Slope }\]it should be noted that the angle with \[\text{x-axis}\], simply means the positive direction of \[\text{x-axis}\] and if a line is perpendicular to \[\text{x-axis}\], then slope cannot be defined, in this condition \[\theta \] will be \[{{90}^{\circ }}\] that means \[\tan {{90}^{\circ }}=\infty \]