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?

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Jim G. Share
Mar 27, 2017

Answer:

#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"#

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dk_ch Share
Mar 27, 2017

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.

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Mar 27, 2017

Answer:

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 :

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