Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:
(i) 2x + 3y = 9.35
(ii) x – \(\frac{y}{5}\)– 10 = 0
(iii) –2x + 3y = 6
(iv) x = 3y
(v) 2x = –5y
(vi) 3x + 2 = 0
(vii) y – 2 = 0
(i) 2x+3y=9.35
⇒ 2x+3y−9.35=0
On comparing this equation with ax+by+c=0,
a=2,b=3 and c=−9.35
(ii) x−\(\frac{y}{5}\)−10=0
On comparing this equation with ax+by+c=0,
a=1,b=−\(\frac{1}{5}\) and c=−10
(iii) −2x+3y=6
⇒ −2x+3y−6=0
On comparing this equation with ax+by+c=0,
a=−2,b=3 and c=−6
(iv) x=3y
⇒ x−3y=0
On comparing this equation with ax+by+c=0, we get,
a=1,b=−3 and c=0
(v) 2x=−5y
⇒ 2x+5y=0
On comparing this equation with ax+by+c=0,
a=2,b=5 and c=0
(vi) 3x+2=0
⇒ 3x+0y+2=0
On comparing this equation with ax+by+c=0,
a=3,b=0 and c=2
(vii) y−2=0
⇒ 0x+y−2=0
On comparing this equation with ax+by+c=0,
a=0,b=1 and c=−2
(viii) 5=2x
⇒ −2x+0y+5=0
On comparing this equation with ax+by+c=0,
a=−2,b=0 and c=5
Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see Fig. 9.27). Prove that ∠ACP = ∠ QCD

ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.14). Show that
(i) ∠A = ∠B
(ii) ∠C = ∠D
(iii) ∆ABC ≅ ∠∆BAD
(iv) diagonal AC = diagonal BD [Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

(i) The kind of person the doctor is (money, possessions)
(ii) The kind of person he wants to be (appearance, ambition)