A B C D 2‌5‌4‌1 (A) x-intercept =5C [Let] y-intercept =3C Line: ‌
x
a
+‌
y
b
=1⇒‌
x
5C
+‌
y
3C
=1. . . (i) Since, it passes through (−4,3). Hence, −‌
4
5C
+‌
3
3C
=1⇒C=‌
1
5
From Eq. (i), ‌‌
x
5×‌
1
5
+‌
y
3×‌
1
5
=1⇒‌
x
1
+‌
5y
3
=1 ⇒‌‌3x+5y=3 (B) Here given, that point P bisect the part intercepted between axes by line, that means P is mid-point of AB. Equation of line in one point form having slope =m is
y+5‌‌=m(x−2) y‌‌=mx−2m−5 . . . (i) ‌ Let ‌A‌‌=(0,−2m−5)⇒B=(2+‌
5
m
,0) We know that, P is mid-point of A and B. ‌ Then, ‌2=‌
0+2+
5
m
2
⇒‌‌4=2+‌
5
m
⇒m=‌
5
2
From Eq. (i), y=‌
5
2
x−2(‌
5
2
)−5⇒5x−2y−20=0 (C) Any line parallel to 2x−3y+5=0 is 2x−3y+λ=0 . . . (i) Since, Eq. (i) passes through (‌
2
5
,0), Hence, 2×‌
2
5
−3×0+λ=0 ⇒‌‌λ=−‌
4
5
From Eq. (i), 2x−3y−‌
4
5
=0 ⇒10x−15y−4=0 (D) Any line perpendicular to 5x+2y+7=0 is 2x−5y+µ=0. . . (i) Since, Eq. (i) passes through (0,‌