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Question: A straight line passes through the point \(\left( {3,2} \right)\) and the portion of this line, inte...

A straight line passes through the point (3,2)\left( {3,2} \right) and the portion of this line, intercepted between the positive axes, is bisected at this point. Find the equation of line?

Explanation

Solution

Wherever we need the equation of a line we have to find only two things: point and a slope. The formula for the calculation of the slope of line is m=y2y1x2x1m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}. Bisection of a line means the division of a line into two equal halves. The line dividing it into two equal halves is called a bisector. We should know various forms of representing a straight line in order to find the equation of the required line.

Complete step by step answer:
We know the two point form of a straight line as (yy1)=(y2y1)(x2x1)(xx1)\left( {y - {y_1}} \right) = \dfrac{{\left( {{y_2} - {y_1}} \right)}}{{\left( {{x_2} - {x_1}} \right)}}\left( {x - {x_1}} \right). So, we need to find the coordinates of two points lying on the required line.Let the line intersect the x axis and y axis at points A(a,0)A\left( {a,0} \right) and B(0,b)B\left( {0,b} \right) respectively. Let P be the point of bisection of the straight line intercepted between the axes.

Now, as we are given that the line is bisected at (3,2)\left( {3,2} \right). So, using the midpoint formula, we have the coordinates of midpoint of a line as (x1+x22,y1+y22)\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right). So, we get the coordinates of the midpoint as (a+02,0+b2)=(a2,b2)\left( {\dfrac{{a + 0}}{2},\dfrac{{0 + b}}{2}} \right) = \left( {\dfrac{a}{2},\dfrac{b}{2}} \right).Hence, we get the coordinates of P as (a2,b2)\left( {\dfrac{a}{2},\dfrac{b}{2}} \right). Also, we are given the coordinates of point of bisection P as (3,2)\left( {3,2} \right).

So, equating both, we get,
a2=3\Rightarrow \dfrac{a}{2} = 3 and b2=2\dfrac{b}{2} = 2
So, finding the values of a and b by solving the equation using transposition method, we get,
a=6\Rightarrow a = 6 and b=4b = 4
Hence, the intercepts of the required line are A(6,0)A\left( {6,0} \right) and B(0,4)B\left( {0,4} \right).

Now, we have the coordinates of three points lying on the line given to us. So, we substitute the coordinates of two of these points into the two point form of a straight line.So, we get,
(y2)=(02)(63)(x3)\left( {y - 2} \right) = \dfrac{{\left( {0 - 2} \right)}}{{\left( {6 - 3} \right)}}\left( {x - 3} \right)
(y2)=(2)3(x3)\Rightarrow \left( {y - 2} \right) = \dfrac{{\left( { - 2} \right)}}{3}\left( {x - 3} \right)
Cross multiplying the terms of the equation, we get,
3y6=2x+6\Rightarrow 3y - 6 = - 2x + 6
Shifting all the constant terms to right side of the equation, we get,
3y+2x=6+6\Rightarrow 3y + 2x = 6 + 6
3y+2x=12\therefore 3y + 2x = 12

So, the equation of the required line passing through the point (3,2)\left( {3,2} \right) and bisected at the same point is 3y+2x=123y + 2x = 12.

Note: We must know the two points form a straight line in order to find the equation of the required line. We also must know the formula for the midpoint of a line segment given the coordinates of the end point. One must have a good grip over concepts of coordinate geometry in order to tackle the problem.