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

4 answers

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