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

Step-by-step explanation:

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MATHS

Points A and B are in the first quadrant; point O is the origin. If the slope of OA is 1,

slope of OB is 7 and OA=OB, then the slope of AB is

December 30, 2019avatar

Sonia Prabakar

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VIDEO EXPLANATION

ANSWER

Let the point A(a,b) and B(c,d)

Slope of OA =

x

2

−x

1

y

2

−y

1

=

a−0

b−0

=1

So, a=b

Slope of OB =

x

2

−x

1

y

2

−y

1

=

c−0

d−0

=7

So, d=7c

Now,

OA=OB

(a−0)

2

+(b−0)

2

=

(c−0)

2

+(d−0)

2

Squaring on both sides

a

2

+b

2

=c

2

+d

2

a

2

+a

2

=c

2

+(7c)

2

2a

2

=50c

2

a

2

=25c

2

a=5c [Only considering positive value since A and B is in the first quadrant]

Slope of AB =

c−a

d−b

=

c−a

7c−a

=

c−5c

7c−5c

=

−4c

2c

=−

2

1

Answer 2

The equation of line which is parallel to the line segment AB and passes through the point P(-3,7) is x + y = 4

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

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

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

$\implies x+y=4$

Hope this answer is helpful,

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