Question

write the equation of a line:
a) parallel to the line 3x+5y=-10
b) parallel to the line 2y-4x=5 and which goes through the point (3,7)
c) perpendicular to the line 8x+6y=-1 and which goes through the point (-2,1)

THANKS!!

Answer 1

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

What is equation of line?

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:  $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$

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= $\frac{5}{2}$ + 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= $\frac{-1}{6} -\frac{4}{3}x$

Hence, m= $-\frac{4}{3}$

We know, if two lines are perpendicular have m and m' as slope then,

m*m'=-1

$\frac{-4}{3} *m'=-1$

 m'= $\frac{3}{4}$

Now, y= $\frac{3}{4}$ x +c

   

Put x= -1 and y=1

then, 1= $\frac{3}{4}$ (-2) + c

         c= $\frac{5}{3}$

So,  y= $\frac{3}{4}$ x + $\frac{5}{2}$

     4y= 3x+ 10

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