19. In an isosceles APQR, PQ = PR. The base QR lies on the x-axis, P lies on the y- axis
and 2x - 3y + 9= 0 is the equation of PQ. Find the equation of the straight line along PR.
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Question :
In an isosceles triangle PQR, PQ = PR. The base QR lies on the axis, P lies on the y-axis and 2 x - 3 y + 9 =0 is the equation of PQ. Find the equation of the straight line along PR.
SOLUTION :
GIVEN :
2 x – 3 y + 9 =0
Point P lies on the y-axis,so x= 0 in the given equation PQ
2 x - 3 y + 9 =0
2(0) - 3 y = - 9
- 3 y = -9
y = (-9)/(-3)
y = 3
Hence,the point P is (0,3)
Point Q lies on the x-axis, so y = 0 in the given equation QR
2 x - 3 y + 9 =0
2 x - 3 (0) = - 9
2 x - 0 = -9
2 x = - 9
x = -9/2
Hence,the point Q is (-9/2 , 0)
PQ = PR , Sides of isosceles ∆ PQR (Given)
So the point R is (9/2,0)
Equation of side PR : P (0 ,3); R (9/2 ,0)
Using two-points formula, the equation of the straight line is :
(y-y₁)/(y₂ - y₁) = (x-x₁)/(x₂ - x₁)
Here, x1= 0, x2= 3, y1=9/2, y2 = 0
(y - 3)/(0 - 3) = [x - 0] /[(9/2) - 0]
(y - 3)/(- 3) = [x ]/[(9/2]
9/2(y-3) = -3x
9/2y - 3×9/2 = -3x
9/2y - 27/2 = -3x
3x + 9/2y - 27/2= 0
3(x +3/2y -9/2)= 0
x +3/2y -9/2= 0
(2x + 3y - 9)/2= 0
2x + 3y - 9 = 0
Hence, 2x + 3y - 9 = 0 is the required equation of the straight line along PR.
HOPE THIS WILL HELP YOU….
In an isosceles triangle PQR, PQ = PR. The base QR lies on the axis, P lies on the y-axis and 2 x - 3 y + 9 =0 is the equation of PQ. Find the equation of the straight line along PR.
SOLUTION :
GIVEN :
2 x – 3 y + 9 =0
Point P lies on the y-axis,so x= 0 in the given equation PQ
2 x - 3 y + 9 =0
2(0) - 3 y = - 9
- 3 y = -9
y = (-9)/(-3)
y = 3
Hence,the point P is (0,3)
Point Q lies on the x-axis, so y = 0 in the given equation QR
2 x - 3 y + 9 =0
2 x - 3 (0) = - 9
2 x - 0 = -9
2 x = - 9
x = -9/2
Hence,the point Q is (-9/2 , 0)
PQ = PR , Sides of isosceles ∆ PQR (Given)
So the point R is (9/2,0)
Equation of side PR : P (0 ,3); R (9/2 ,0)
Using two-points formula, the equation of the straight line is :
(y-y₁)/(y₂ - y₁) = (x-x₁)/(x₂ - x₁)
Here, x1= 0, x2= 3, y1=9/2, y2 = 0
(y - 3)/(0 - 3) = [x - 0] /[(9/2) - 0]
(y - 3)/(- 3) = [x ]/[(9/2]
9/2(y-3) = -3x
9/2y - 3×9/2 = -3x
9/2y - 27/2 = -3x
3x + 9/2y - 27/2= 0
3(x +3/2y -9/2)= 0
x +3/2y -9/2= 0
(2x + 3y - 9)/2= 0
2x + 3y - 9 = 0
Hence, 2x + 3y - 9 = 0 is the required equation of the straight line along PR.
HOPE THIS WILL HELP YOU….
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It is given that ,
∆PQR is an isosceles triangle.
PQ = PR , base QR lies on the
x-axis ,P lies on the y-axis.
and
equation of PQ is 2x - 3y + 9 = 0
i ) Convert PQ in intercepts form ,
2x - 3y = -9
divide each term with -9 , we get
=> 2x/(-9 ) + (-3y)/(-9) = ( -9)/(-9)
=> x/(-9/2) + y/(-9/-3) = 1
=> x/(-9/2) + y/3 = 1
x-intercept = a = -9/2 ,
y-intercept = b = 3 ,
ii ) R lies on X-axis ,
coordinates of R = ( 9/2 , 0 ) =(x1,y1)
P lies on y - axis ,
coordinates of P = ( 0, 3 )=(x2,y2)
Equation of PR ,
y-y1 = [ (y2-y1)/(x2-x1) ] ( x - x1 )
=> y -0 = [(3-0)/(0-9/2)] ( x-9/2)
=> y = (-2/3)[(2x-9)/2]
=> y = (-1/3)(2x-9)
=> 3y = -(2x - 9)
=>2x + 3y - 9 = 0
Therefore ,
equation of PR is
2x + 3y - 9 = 0
••••
=>
∆PQR is an isosceles triangle.
PQ = PR , base QR lies on the
x-axis ,P lies on the y-axis.
and
equation of PQ is 2x - 3y + 9 = 0
i ) Convert PQ in intercepts form ,
2x - 3y = -9
divide each term with -9 , we get
=> 2x/(-9 ) + (-3y)/(-9) = ( -9)/(-9)
=> x/(-9/2) + y/(-9/-3) = 1
=> x/(-9/2) + y/3 = 1
x-intercept = a = -9/2 ,
y-intercept = b = 3 ,
ii ) R lies on X-axis ,
coordinates of R = ( 9/2 , 0 ) =(x1,y1)
P lies on y - axis ,
coordinates of P = ( 0, 3 )=(x2,y2)
Equation of PR ,
y-y1 = [ (y2-y1)/(x2-x1) ] ( x - x1 )
=> y -0 = [(3-0)/(0-9/2)] ( x-9/2)
=> y = (-2/3)[(2x-9)/2]
=> y = (-1/3)(2x-9)
=> 3y = -(2x - 9)
=>2x + 3y - 9 = 0
Therefore ,
equation of PR is
2x + 3y - 9 = 0
••••
=>
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