The line 4x-3y+12=0 meets x-axis at A.Write the co-ordinates of A.
Determine the equation for the line through A and perpendicular to 4x-3y+12=0.
Answers
Answered by
16
By using intercept form we can know a co ordinates
4x/-12-+3y/12=1
X/-12/4 +y/12/3=1
X/-3+y/4=1
A co ordinates (-3,0)
Now slope of line is -a/b if of line is ax+by +c=0
By comparing we get slope for line 4x-3y+12=0
Slope is 4/3
For the slope of perpendicular m1m2 =-1
M2=-1/m1 so slope is -3/4
By line equation
(Y-0)=-3/4 (X+3)
4y=-3x-9
3x+4y+9=0 req line eq.
4x/-12-+3y/12=1
X/-12/4 +y/12/3=1
X/-3+y/4=1
A co ordinates (-3,0)
Now slope of line is -a/b if of line is ax+by +c=0
By comparing we get slope for line 4x-3y+12=0
Slope is 4/3
For the slope of perpendicular m1m2 =-1
M2=-1/m1 so slope is -3/4
By line equation
(Y-0)=-3/4 (X+3)
4y=-3x-9
3x+4y+9=0 req line eq.
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I can even find co ordinates by substituting x=x and y =I since point is (x,0) on x axis we get 4x=-12 and c =-3
X=-3
Answered by
6
since the pointA is on the x axis the y coordinate of the point will be 0. put y=0 in the equation of the line and you will get the x coordinate which is -3 A(-3,0).
slope of the line 4x-3y+12=0 is 4/3so the slope of the line perpendicular to it will be -3/4.
so we will use the formula y-y1=m(x-x1).
y-0=(-3/4)(x+3)
4y=-3x-9
therfore the equation becomes 3x +4y+9=0.
slope of the line 4x-3y+12=0 is 4/3so the slope of the line perpendicular to it will be -3/4.
so we will use the formula y-y1=m(x-x1).
y-0=(-3/4)(x+3)
4y=-3x-9
therfore the equation becomes 3x +4y+9=0.
sorry it's wrong
Yup it's wrong...simple mistake...edit it
Co ordinordinates x=x and y is 0
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