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Question: If the line \( 7y - x = 25 \) cuts the curve \( {x^2} + {y^2} = 25 \) at the points A and B. Find th...

If the line 7y−x=257y - x = 25 cuts the curve x2+y2=25{x^2} + {y^2} = 25 at the points A and B. Find the equation of the perpendicular bisector of the line AB.

Explanation

Solution

Hint : We are given the equation of a line that cuts a curve, so their point of intersection will satisfy both the equations, so we can find out the coordinates of the points A and B using this approach. After finding the coordinates, we have to find out the equation of the perpendicular bisector of the line joining these coordinates. As we know the coordinates of the points A and B, we can find out its slope and thus get the slope of the perpendicular bisector and by finding out one passing point of the perpendicular bisector by midpoint formula, we can easily find out its equation.

Complete step-by-step answer :
Equation of the line is 7y−x=257y - x = 25 , it can be rewritten as –
x=7y−25x = 7y - 25
Equation of the curve is x2+y2=25{x^2} + {y^2} = 25 , putting the value of x in this equation, we get –
(7y−25)2+y2=25 49y2+625−350y+y2=25 50y2−350y+600=0 50(y2−7y+12)=0 ⇒y2−7y+12=0   {(7y - 25)^2} + {y^2} = 25 \\\ 49{y^2} + 625 - 350y + {y^2} = 25 \\\ 50{y^2} - 350y + 600 = 0 \\\ 50({y^2} - 7y + 12) = 0 \\\ \Rightarrow {y^2} - 7y + 12 = 0 \;
Now, the obtained equation is a quadratic polynomial equation that can be solved by factorization as –

y2−4y−3y+12=0 y(y−4)−3(y−4)=0 (y−3)(y−4)=0 ⇒y=3, y=4   {y^2} - 4y - 3y + 12 = 0 \\\ y(y - 4) - 3(y - 4) = 0 \\\ (y - 3)(y - 4) = 0 \\\ \Rightarrow y = 3,\,y = 4 \;

x=7(3)−25, x=7(4)−25 ⇒x=−4, x=3   x = 7(3) - 25,\,x = 7(4) - 25 \\\ \Rightarrow x = - 4,\,x = 3 \;
So the coordinates of the point of intersection of the given line and the curve are A(−4,3)A( - 4,3) and B(3,4)B(3,4) .
We have to find the equation of the perpendicular bisector of the line AB, so the coordinates of the point at which the perpendicular cuts the line will be –
(x,y)=(−4+32,3+42) ⇒(x,y)=(−12,72)   (x,y) = (\dfrac{{ - 4 + 3}}{2},\dfrac{{3 + 4}}{2}) \\\ \Rightarrow (x,y) = (\dfrac{{ - 1}}{2},\dfrac{7}{2}) \;
The slope of line AB is –
m1=y2−y1x2−x1=4−33−(−4) ⇒m1=17  {m_1} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \dfrac{{4 - 3}}{{3 - ( - 4)}} \\\ \Rightarrow {m_1} = \dfrac{1}{7} \\\
So, the slope of the perpendicular bisector is –
m2=−1m1=−117 ⇒m2=−7   {m_2} = \dfrac{{ - 1}}{{{m_1}}} = \dfrac{{ - 1}}{{\dfrac{1}{7}}} \\\ \Rightarrow {m_2} = - 7 \;
Now, we know the slope and the passing point of the perpendicular bisector, equation of the perpendicular bisector is –
y−y1=m2(x−x1) y−72=−7[x−(−12)] 2y−72=−14x−72 2y−7=−14x−7 ⇒14x+2y=0 ⇒7x+y=0   y - {y_1} = {m_2}(x - {x_1}) \\\ y - \dfrac{7}{2} = - 7[x - (\dfrac{{ - 1}}{2})] \\\ \dfrac{{2y - 7}}{2} = \dfrac{{ - 14x - 7}}{2} \\\ 2y - 7 = - 14x - 7 \\\ \Rightarrow 14x + 2y = 0 \\\ \Rightarrow 7x + y = 0 \;
Hence the equation of the perpendicular bisector is 7x+y=07x + y = 0 .
So, the correct answer is “ 7x+y=07x + y = 0 ”.

Note : The perpendicular bisects the given line AB that is it cuts the line in two halves that’s why we find the coordinates using the midpoint formula. We know that the product of the slope of two perpendicular lines is -1 so the slope of a line perpendicular to the given line of the slope m1{m_1} is given as −1m1\dfrac{{ - 1}}{{{m_1}}} .