12x2+7xy−12y2=0 ‌12x2+16xy−9xy−12y2=0 ‌4x(3x+4y)−3y(3x+4y)=0 ‌(4x−3y)(3x+4y)=0 ‌4x−3y=0 and 3x+4y=0 Slope of the two lines are ‌
4
3
and ‌
−3
4
, respectively. The two lines are perpendicular. ‌12x2+7xy−12y2−x+7y−1=0 ‌12x2+x(7y−1)−12y2+7y−1=0 ‌x=−‌
b+√b2−4ac
2a
‌x=−‌
(7y−1)±√(7y−1)2−4×12(−12y2+7y−1)
2×12
‌x=‌
1−7y±√49y2+1−14y−336y+48+576y2
24
‌⇒24x=1−7y±√625y2−350y+49 ‌⇒24x=1−7y±(25y−7) ‌⇒24x=1−7y+25y−7 or 24x=1−7y−25y+7 ‌⇒24x−18y+6=0 or 24x+32y−8=0 m3‌=‌
24
18
and m4=‌
−24
32
⇒m3‌=‌
4
3
and m4=‌
−3
4
Again, m3×m4=−1, the two lines are perpendicular. The two lines 3x+4y=0 and 4x−3y=0 intersect at a point (0,0). d1, and d2 are the distance of a point (0,0) to the line 24x−18y+6=0 and 24x+32y−8=0
‌d1=‌
6
√(24)2+(18)2
,d2=‌
8
√(24)2+(32)2
‌d1=‌
6
30
=‌
1
5
,d2=‌
8
40
=‌
1
5
∴ The given figure is a square of side 1∕5 unit. Area of square =‌