Question

In the figure the equation of ab is x+2y=1, p is (3,1) and angle PRA=angle QRB.if the slope of pr is 1 find the equation of the line QR​

Answer 1

Answer:

QR is y = 7x - 12

This answer is absolutely correct I have verified it with two ways.

I'll upload the shortest answer.

Step-by-step explanation:

Step 1

Finding equation of PR:

Slope of PR = 1

PR is,

( y - y1 ) = m ( x - x1 )

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

=> y - 1 = x - 3

=> y = x - 2

Therefore, PR is y = x - 2

Step 2

Finding R:

Since both lines PR and AB intersect at R.

Therefore, y = x - 2 = ( 1 - x ) / 2

[ Since, PR is y = x - 2 and

AB is x + 2y = 1 ]

=> 2x - 4 = 1 - x

=> 3x = 5

=> x = 5 / 3

Now,

y = x - 2 or y = ( 1 - x ) / 2

=> y = ( 3 / 5 ) - 2 or y = ( 1 - (3 / 5) ) / 2

=> y = (- 1) / 3 or y = (- 2 / 3) / 2

=> y = - ( 1 / 3 )

Therefore,

R is ( 5/3 , -1/3)

Step 3

Construct a line passing through P and Q parallel to AB and a perpendicular from R on PQ

In ∆ PXR and ∆ QXR,

• angel PXR and QXR are equal [ Each 90° ]

• XR = XR [ Common ]

• angle XRP and XRQ are equal [ angel XRA and XRB are 90° and angel PRA and QRB are equal as given ]

Therefore, ∆ PXR ≅ ∆ QXR [ By ASA axiom ]

Therefore, PX = QX [ CPCTC ]

Step 4

Finding equation of RX and PQ:

Slope of PQ = slope of AB [ AB || PQ ]

= -1/2

Now,

PQ is,

( y - y1 ) = m ( x - x1 )

=> ( y - 1 ) = -1/2 ( x - 3 )

=> 2( y - 1 ) = 3 - x

=> 2y - 2 = 3 - x

=> 2y = 5 - x

=> y = ( 5 - x ) / 2

Since, RX ⊥ PQ

Therefore, slope of RX = 2 [ m1 = 1/m2 ]

RX is,

( y - y1 ) = m ( x - x1 )

=>

I'll upload soon it's a bit long...