Question

Suppose A and B are points in the first quadrant such that the slopes of the line segments. OA and OB are 3 and 1/3 respectively, where O is the origin. If OA = OB , then what is the equation of line which is parallel to the line segment AB and passes through the point P(3,7)?

Answer 1

The equation of line which is parallel to the line segment AB and passes through the point P(3,7) is x + y = 10

Step-by-step explanation:

Given,

the slopes of the line segment OA and OB = 3, 1/3 respectively

Since the line segments pass through origin

The equation of line OA is

$y=3x$

$\implies 3x-y=0$

And the equation of line OB is

$y=\frac{1}{3}x$

$\implies x-3y=0$

We know that if two lines are $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$, then the equation of their angle bisector is given by

$\boxed{\frac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}=\pm\frac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}}$

Where we take negative sign if $a_1a_2+b_1b_2>0$

And positive sign if $a_1a_2+b_1b_2<0$

Thus the equation of angle bisector of angle AOB

$\frac{3x-y}{\sqrt{3^2+1^2}}=\pm\frac{x-3y}{\sqrt{1^2+3^2}}$

$\implies 3x-y=-(x-3y)$ (∵$a_1a_2+b_1b_2>0$, ∴ taking negative sign)

$\implies 3x-y=-x+3y$

$\implies 4x-4y=0$

$\implies y=x$

This is the equation of line segment OC

Slope of OC, $m_1=1$

But given that OA = OB

Therefore, ΔOAB is an isosceles triangle

Therefore, the angle bisector OC will be perpendicular to AB

Thus, if the slope of AB is $m_2$

Then,

$m_1m_2=-1$

$\implies 1\times m_2=-1$

$\implies m_2=-1$

Since line passing through point P (3, 7) is parallel to AB

Therefore, the slope of line passing through P and parallel to AB will also be $m=-1$

The equation of the line will be

$y-7=-1(x-3)$

$\implies y-7=-x+3$

$\implies x+y=10$

Hope this answer is helpful,

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Answer 2

x + y  = 10  is the line parallel to the line segment AB and passes through the point P(3,7)

Step-by-step explanation:

OA  Slope 3

Passes throught origin

y - 0 = 3(x - 0)

=> y = 3x

Let say A  =  (a  , 3a)

OB Slope  =  1/3

=> y - 0 = (1/3)(x - 0)

=> y  = x/3

Let say B = ( 3b , b)

OA  = √a² + (3a)²  =  √10a

OB = √10b

OA = OB

=> a  =  b      ( as both are in 1st Quadrant so both would be + ve)

A = ( a , 3a)

B = (3a , a)

Slope of AB  =  (  a - 3a)/(3a - a)  = - 1

Passes through point ( 3 , 7)

=> y - 7  = -(x  - 3)

=> y - 7 = -x  + 3

=> x + y  = 10

x + y  = 10  is the line parallel to the line segment AB and passes through the point P(3,7)

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