The slope of the line determined by the equation 3 − 2 = −4 is 3
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Equation of line with given slope
4
3
and from point A(3,2)
y−2=
4
3
(x−3)
4y−8=3x−9
3x=1+4y------(1)
Let Point B(x
1
,y
1
)
The distance between A and B is 5 Hence by distance formula
(x
1
−3)
2
+(y
1
−2)
2
=25
x
1
2
+9−6x
1
+y
1
2
+4−4y
1
=25
x
1
2
+y
1
2
−6x
1
−4y
1
−12=
Point B lies on line 3x=1+4y
Hence x
1
=
3
4y
1
+1
(
3
4y
1
+1
)
2
+y
1
2
−6(
3
4y
1
+1
)−4y
1
−12=0
9
16y
1
2
+1+8y
1
+y
1
2
−8y
1
−2−4y
1
−12=0
16y
1
2
+1+8y
1
+9y
1
2
−72y
1
−18−36y
1
−108=0
25y
1
2
−100y
1
−125=0
y
1
2
−4y
1
−5=0
y
1
2
−5y
1
+y
1
−5=0
(y
1
−5)(y
1
+1)=0
⇒y
1
=−1,5
From equation (1)
3x
1
=1+4y
1
3x
1
=1−4 and 3x
1
=1+20
x
1
=−1 and x
1
=7
Point B(7,5) and (−1,−1)
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