the sides of a triangle are 3x+2y-6=0, 2x-3y+6=0, x+2y+2=0 ;p(0,b) is a point on y axis. if p lies on the triangle or inside the triangle then the range of b is
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The equation of two sides of a △ are 3x−2y+6=04x+5y−20=0 resp. The orthocentre of △ is (1,1) find equation of third side which lies on the point (
2
35
,−10) ?
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Answer
Let equation of AB be : 3x−2y+6=0 ___(1)
Let equation of AC be : 4x+5y−20=0 __(ii)
3x−2y+6=0⇒x=
3
2y−6
___ from (i)
replacing value of x in (ii)
4×
3
(2y−6)
+5y−20=0
⇒8y−24+15y−60=0
⇒23y=84⇒y=
23
84
∴(x,y)=(
23
10
,
23
89
)
now, x=
3
2×
23
89
−6
=
3×23
168−138
=
3×23
30
=
23
10
∴ equation of alt. AP ⇒
x−1
y−1
=
23
10
−1
23
84
−1
⇒
x−1
y−1
=
−13
61
⇒−13y+13=61x−61
⇒61x+13y−74=0
∵BC⊥AP
∴m(BC)=
m
(AP)
−1
now, for m
AP
equation of AP ⇒61x+13y−74=0
⇒13y=74−61x⇒y=
13
−61
x+
13
74
∴m
BC
=
−
13
61
−1
=
61
13
∴ equation of BC⇒y=
61
13
x+c
[∵y=mx+c]
⇒61y−13x=c
As, BC lies on (
2
35
,10)
∴ equation of BC⇒61⋅(−10)−13⋅
2
35
=c=
2
−1675
∴ equation of BC becomes ⇒61y−13x+
2
1675
=0
⇒13x−61y−
2
1675
=0
Answer:
here is your ans.
Step-by-step explanation:
Let equation of AB be : 3x−2y+6=0 ___(1)
Let equation of AC be : 4x+5y−20=0 __(ii)
3x−2y+6=0⇒x=
32y−6
___ from (i)
replacing value of x in (ii)
4× 3(2y−6)
+5y−20=0
⇒8y−24+15y−60=0
⇒23y=84⇒y= 2384
∴(x,y)=( 2310 2389 )
now, x= 32× 2384 −6
= 3×23168−138
= 3×2330
= 2310
∴ equation of alt. AP ⇒
x−1y−1 = 2310 −12384 −1
⇒ x−1y−1
= −1361
⇒−13y+13=61x−61
⇒61x+13y−74=0
∵BC⊥AP
m(BC)= m (AP)−1
now, for m AP
equation of AP ⇒61x+13y−74=0
⇒13y=74−61x⇒y= 13−61 x+ 1374
∴m BC = − 1361−1
= 6113
∴ equation of BC⇒y= 6113
x+c [∵y=mx+c]
⇒61y−13x=c
As, BC lies on ( 235 ,10)
∴ equation of BC⇒61⋅(−10)−13⋅ 235
=c= 2−1675
∴ equation of BC becomes ⇒61y−13x+ 21675
=0⇒13x−61y− 21675
=0solution expand