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?

Question

Jayden Jackson · Accepted Answer

Contd.

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 :

Sadie Allen · Answer

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.

Caroline Aucoin · Answer

#y=2x-6#

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"#

Autumn Armstrong · Answer

undefined First, we need to find the point of intersection of the given lines. We solve the system of equations formed by the two lines:

7x - 3y - 19 = 0
3x + 2y + 5 = 0

By solving this system, we find the coordinates of the point of intersection.

Once we have the coordinates of the point of intersection, we can use the fact that the line we're looking for is parallel to the line with the equation y = 2x + 1. Since parallel lines have the same slope, we know that the slope of the line we're looking for is also 2.

Now, we can use the point-slope form of a linear equation, which is y - y₁ = m(x - x₁), where (x₁, y₁) is a point on the line and m is the slope.

Substituting the coordinates of the point of intersection and the slope into the point-slope form, we can find the equation of the line.