a) Line parallel to the line 3x+5y=-10 is -6x-10y=9
b) Line parallel to the line 2y-4x=5 and which goes through the point
(3,7) is y= 2x-1.
c) Line perpendicular to the line 8x+6y=-1 and which goes through the
point (-2,1) is 4y= 3x+ 10
The standard form of equation of a line is ax + by + c = 0. Here a, b, are the coefficients, x, y are the variables, and c is the constant term. It is an equation of degree one, with variables x and y.
a) Equation parallel to 3x+5y=-10 is
-6x-10y=9
So, it satisfies:
b)given equation into the form of y=mx+c,
2y-4x=5
we can find the gradient of the line.
2y= 5+4x
y= + 2x
Hence, the gradient is 2.
As, Parallel lines have the same gradient
Thus the line would also have a gradient of 2.
Substitute m=2 into the equation:
y= 2x +c
To find the value of c, substitute a pair of coordinates.
When x=3, y=7,
7= 2(3) +c
7= 6 +c
c=6-7
c= -1
So, equation of line be: y= 2x-1.
c) Given: 8x+6y=-1
6y= -1-8x
y=
Hence, m=
We know, if two lines are perpendicular have m and m' as slope then,
m*m'=-1
m'=
Now, y= x +c
Put x= -1 and y=1
then, 1= (-2) + c
c=
So, y= x +
4y= 3x+ 10
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