Question

1. The equation, in general form, of the line that passes through the point (3,4) and is perpendicular to the line 5 x + 5 y + 9 = 0 is Ax + By + C = 0, where a=? b=? c=?

2.The equation of the line that goes through the points ( -9 ,-6 ) and ( -6 ,-3 ) can be written in general form A x + B y + C = 0 where B=?

Answer 1

1.
The line perpendicular to the given line
Ax + By + C = 0
is
Bx - Ay + D = 0
where D has to be found from an intercept or a given point through which the new line passes.
For the given line
5x + 5y + 9 = 0
the required line is
5x - 5y + D = 0
Since it passes through (3,4)
D can be found by substituting x=3 and y=4
5*3 -5*4 + D = 0
D = 5
Thus the required line is
5x -5y + 5 = 0
check: 5(3)-5(4)+5 =15-20+5=0 OK

2.
A line passing through two given points P1(x1,y1) and P2(x2,y2) is given by
(y2-y1)(x-x1) - (x2-x1)(y-y1) =0
substitute P1(-9,-6), P2(-6,-3) to get
(-3-(-6)(x-(-9)) - (-6-(-9))(y-(-6))=0
or
x-y+3=0
after dividing both sides of the equal sign by 3.
A=1, B=-1, C=3.
Check
-9-(-6)+3=0
-6-(-3)+3=0
OK.