Question

How to find the equation of the line which passes through the point of intersection of the lines 7x − 3y − 19 = 0 and 3x + 2y + 5 = 0, give that the line is parallel to the line with the equation y = 2x + 1?

Answer 1

Contd.

Explanation:

We will use the following well-known

Result : The eqn. of a line `l` passing through the pt. of

intersection of the lines `l_1 : a_1+b_1+c_1=0 and l_2 : a_2x+b_2y+c_2=0` is of

the form

Result :

Answer 2

Let equation of any straight line passing through the point of intersection of two given straight line be

`k(7x-3y-19)+(3x+2y+5)=0`

#=>(7k+3)x+(2-3k)y+(5-19k)=0.. [1]#

If the straight be parallel to the straight line `2x-y+1=0`

then

`(7k+3)/2=(2-3k)/(-1)`

`=>7k+3=-4+6k`

`=>k=-7`

Inserting the value of k in [1]

`(7*(-7)+3)x+(2-3*(-7))y+(5-19*(-7))=0`

`=>-46x+23y+138=0`

`=>2x-y-6=0`

This is the equation of the required straight line.

Answer 3

`y=2x-6`

Explanation:

The first step is to find the point of intersection of the 2 lines.

`"Using the "color(blue)"elimination method"`

That is we attempt to eliminate the x or y term from the equations leaving us with an equation in 1 variable which we can solve.

Labelling the equations.

`color(red)(7x)color(magenta)(-3y)-19=0to(1)`

`color(red)(3x)color(magenta)(+2y)+5=0to(2)`

`"Note:" color(magenta)(-3y)xx2=color(magenta)(-6y)" and "color(magenta)(2y)xx3=color(magenta)(6y)`

That is the y terms have the same coefficient but with opposing signs. Hence summing them will result in their elimination.

`(1)xx2: 14x-6y-38=0to(3)`

`(2)xx3: 9x+6y+15=0to(4)`

`(3)+(4)" term by term"`

`rArr23x+0y-23=0larrcolor(blue)" equation in one variable"`

`rArr23x=23rArrx=1larrcolor(blue)"value for x"`

Substitute this value into either of ( 1 ) or ( 2 ) and solve for y

`"Substitute " x=1" in " (2)`

`rArr(3xx1)+2y+5=0`

`rArr8+2y=0rArr2y=-8rArry=-4larrcolor(blue)"value for y"`

`color(blue)"As a check"`

Substitute these values into ( 1 )

`(7xx1)-(3xx-4)-19=7+12-19=0to" true"`

`rArr(1,-4)color(red)" is the point of intersection"`

`color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(" parallel lines have equal slopes")color(white)(2/2)|)))`

`y=2x+1" is in " color(blue)"slope-intercept form"`

`rArr"slope " =m=2`

Expressing the required equation in `color(blue)" point-slope form"`

`y-y_1=m(x-x_1)" with " m=2" and " (x_1,y_1)=(1,-4)`

`rArry+4=2(x-1)larrcolor(red)" in point-slope form"`

distribute and simplify.

`y+4=2x-2`

`rArry=2x-6larrcolor(red)" in slope-intercept form"`