ax +by=c bx-ay=o/ solve
Answers
Answer࿐
Given,
3x+2y = -4 ----------(1)
2x+5y = 1 -----------(2)
¶ Find Point of Intersection of these 2 lines
Do 2×(1) - 3×(2)
6x+4y = -8 ----------(3)
6x+15y = 3 ---------(4)
-----------------------
-11y = -11
=> y = 1
Substitute in (1)
3x+2(1) = -4
=> 3x = -4-2
=> 3x = -6
=> x = -2
•°• Point of Intersection of the lines 3x+2y+4= 0 & 2x+5y-1= 0 is (-2,1)
¶ By using the Slope-Intercept form, find the form of equation of line passing through the point (-2,1)
y = mx + c
substitute x = -2 & y = 1
=> 1 = -2m + c
=> c = 1 + 2m
•°• The Required Equations of straight line is of form :
y = mx + 1 + 2m -----------(5)
¶ The perpendicular distance (or simply distance) 'd' of a point P(x1,y1) from Ax+By+C = 0 is given by
Given,
(x1,y1) = (-2,1) & d = 2
=> (4m+2)² = 2(m² + 1)
=> 16m² + 16m + 4 = 2m² + 2
=> 14m² + 16m + 2 = 0
=> 7m² + 8m + 1 = 0
Factorise the equation
=> 7m² + 7m + m + 1 = 0
=> 7m(m+1) + 1(m+1) = 0
=> (m+1)(7m+1) = 0
=> m = -1 and m = -1/7
Now Substitute m = -1 in (5)
=> y = (-1)x+1+2(-1)
=> y = -x + 1 - 2
=> y = -1 - x
(or)
=> -x - y -1 = 0
=> x + y + 1 = 0 ------------(6)
Substitute m = -1/7 in (5)
=> y = (-1/7)x + 1 + 2(-1/7)
=> y = -x/7 + (7-2)/7
=> y = (-x+5)/7
=> 7y = -x+5
or
=> -x - 7y + 5 = 0
=> x + 7y - 5 = 0 ----------(7)
•°• The Required equation of straight lines are :
x + y + 1 = 0 & x + 7y - 5 = 0
✌✌✌✌✌✌✌
hope that u r helpful
please follow thank and brinlylist please
Answer:
Given equations are,
ax+by=c ....(1)
and bx+ay=1+c ....(2)
Multiply equation (1) by a and equation (2) by b, we get
a
2
x+aby=ac ....(3)
and b
2
x+aby=b+bc ....(4)
Subtract equations (3) and (4),
x(a
2
−b
2
)=ac−b−bc
⇒x=
a
2
−b
2
ac−b−bc
Put this value in equation (1), we get
a(
a
2
−b
2
ac−b−bc
)+by=c
⇒a
2
c−ab−abc−a
2
c+b
2
c=−by(a
2
−b
2
)
⇒b(bc−ab−a)=−by(a
2
−b
2
)
⇒y=
b
2
−a
2
bc−a−ac
Step-by-step explanation:
hope u understand please make me brainliest