{SOLVE IT }
Find the equation of the straight line perpendicular to the line 9x+3y=0 and passing through the point of intersection of the lines x+8y-1=0 and x-3y+4=0.
Answers
Answer:
x + 3y + 2 = 0
Step-by-step explanation:
x+6y-1=0 .. (1)
x-3y+4=0 .. (2)
y = 5/11 = 0.454
substitute y in 2 we get
x = -2.639
Point of intersection of above both lines is (0.454, -2.639)
Given straight line
9x + 3y = 0
Slope of line m1 = 3
Slope of line m2 = - 1/3
So, the equation of line passes through intersection point
y + 1 = -1/3(x + 1)
3y + 3 = -x + 1
x + 3y + 2 = 0
Answer:
let the required line be y=mx+c
its perpendicular to 4x−2y=3
Slope of 4x−2y=3 ⇒ 4x−2y=3⇒ m
1
=
2
4
=2
m
1
×m=−1
2×m=−1
m=−
2
1
x+c
∴ y=−
2
1
x+c
Passes through intersection of 5x−8y+23=0 and 7x+6y−71=0
3(5x−8y+23)=0⇒ 15x−24y+69=0
64(7x+6y−71)=0 ⇒ 28x+24y−284=0
---------------------------------------------------------------------------------------
43x=284−69
43x=215
x=
43
215
x=5
5(5)−8y+23=0
8y=48⇒y=6
meeting point P=(5,6)
P passes through y=mx+c
⇒ 6=
2
1
(5)+c
c=6+
2
5
=
2
17
∴ y=−
2
x
+
2
17
⇒ 2y+x−17=0
Equation of the line is 2y+x−17=0
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