Show that a straight line (a+2b)x+(a-b)y=a-7b for different values of a and b passes through a fixed point and aslo - Maths - Straight Lines

Question

Show that a straight line (a+2b)x+(a-b)y=a-7b for different
values of a and b passes through a fixed point and aslo find the
fixed point?
If the point (a,0)lies on the the line containing the the
points (at12,2at1)(at22,2at2) Prove that
t18t2=-1?
Find the image of the point (1,2) in the line x-3y+4=0 assuming
the line to be a plane mirror

Niharika Khare · Accepted Answer

1)
Given straight line equation is
:(a+2b)x+(a-b)y=a-7bax+2bx+ay-by=a-7b(x+y)a+(2x-y)b=a-7bcomparing
the coefficients of a and b on both the sides of above eq
x+y=12x-y=-7Solving the above two eq s for x and y, we get:x=-2 and
y=3Thus, the above given straight line will always pass through a
fixed point, i e (-2,3)

2)
Equation of a straight line passing through the points (at12,2at1)
and (at22,2at2) is given
by:y-y1x-x1=y2-y1x2-x1y-2at1x-at12=2at2-2at1at22-at12y-2at1x-at12=2t2-2t1t22-t12=2(t2-t1)t22-t12=2t2+t1y-2at1x-at12=2t2+t1SInce
(a,0) lies on the above line, it will satisfy the above
equation:0-2at1a-at12=2t2+t1-2t11-t12=2t2+t1-t11-t12=1t2+t1-t1(t2+t1)=1-t12-t1t2-t12=1-t12t1t2=-1

3)
Given straight line is : x-3y+4=0
Now, image of point (1,2) will lie on the line perpendicular to
â€‹x-3y+4=0 and passing through (1,2)
<img alt="" src=
"https://s3mn%20mnimgs%20com/img/shared/content_ck_images/images/MN_Q3(1)%20png"
style="width: 451px; height: 255px;">
Now, x-3y+4=0⇒3y=x+4⇒y=x3+43⇒Slope of this line is
13Slope of the line perpendicular to x-3y+4= is =-113=-3Now, eq of
the line with slope -3 and passing through (1,2)
is:y-2x-1=-3-3x+3=y-23x+y-5=0Let the point of intersection of the
two perpendicular lines be M, then solving x-3y+4=03x+y-5=0we get,
x=1110, y=1710 as point MNow, let Q'(a,b) be the image of point
Q(1,2), then M1110,1710 will be the mid point of Q and Q' such
that:1+a2=1110 ⇒1+a=115⇒a=115-1=652+b2=1710
⇒2+b=175⇒a=175-2=75Thus, the image of point (1,2) is
65,75