write the general form of the equation of the lines throught the point
(a) parallel to the given line and (b) perpendicular to the given point
point- (2,1)
line- 4x-2y=3
Given a line
L : Ax+By+C=0 and a point P1(x1,y1),
The line passing through P1 and parallel to L is given by:
A(x-x1)+B(y-y1)=0
The line passing through P1 and perpendicular to L is given by:
B(x-x1)-A(y-y1)=0
Post if you need more help.
yes i need help can you solve it out then i will ask you questions from then
I will give you an example how to apply those formulas, and would appreciate if you could show me the solution to the posted question. I will be glad to check the answer for you.
Here we go.
Given
L : 4x + 3y = 2, and
P1 : (5,2)
a. Find the line L1 passing through P1 and parallel to L, and
b. L2 passing through P1 and perpendicular to L.
a. Line L1 parallel to L
L : 4x + 3y -2 = 0, P1(5,2)
A=4, B=3, C= -2, x1=5, y1=2
L1 : A(x-x1)+B(y-y1)=0
L1 : 4(x-5)+3(y-2)=0
L1 : 4x -20 + 3y -6 =0
L1 : 4x + 3y - 26 =0
check: substitute x=5, y=2
4(5)+3(2)-26 = 20+6-26=0
Therefore L1 passes through P1.
Slope: m = -4/3
b. Line L2 perpendicular to L
L : 4x + 3y -2 = 0, P1(5,2)
A=4, B=3, C= -2, x1=5, y1=2
L2 : B(x-x1)-A(y-y1)=0
L2 : 3(x-5) - 4(y-2) = 0
L2 : 3x -15 - 4y + 8 = 0
L2 : 3x - 4y - 7 = 0
check: substitute x=5, y=2
L2 : 3(5)-4*2-7 = 15-8-7=0 L2 passes through P1.
ok this is a problem similar to it correct
To find the general form of the equation of a line that passes through a given point, you need to consider its slope.
(a) To find the equation of a line parallel to the given line, you need to identify their slopes. The slope of a line can be determined by rearranging the equation into slope-intercept form (y = mx + b), where m is the slope. For the given line, 4x - 2y = 3, rearranging it to slope-intercept form gives:
-2y = -4x + 3
y = 2x - 3/2
Comparing this equation with y = mx + b, we can see that the slope (m) of the given line is 2.
Parallel lines have the same slope, so any line parallel to this given line would also have a slope of 2. Using the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) represents the given point, (2, 1), and m represents the slope (in this case, m = 2), we can substitute these values to find the equation:
y - 1 = 2(x - 2)
y - 1 = 2x - 4
y = 2x - 3
Therefore, the general form of the equation of a line parallel to the given line and passing through the point (2, 1) is 2x - y = 3.
(b) To find the equation of a line perpendicular to the given line, you need to find the negative reciprocal of its slope. The negative reciprocal of a slope can be found by taking the negative reciprocal of its coefficient. So, for the given line, the slope is 2. The negative reciprocal of 2 is -1/2.
Using the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) represents the given point, (2, 1), and m represents the slope (in this case, m = -1/2), we can substitute these values to find the equation:
y - 1 = (-1/2)(x - 2)
y - 1 = (-1/2)x + 1
y = (-1/2)x + 2
Therefore, the general form of the equation of a line perpendicular to the given line and passing through the point (2, 1) is x + 2y = 4.