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Question : 45 of 160
Marks:
+1,
-0
Solution:
Given,
ABCD is a rectangle
Slope of
AB=‌=−3=m1....(i)
Slope of
BC=‌=m2 ....(ii)
Now
‌‌AB⟂BC⇒m1m2=−1.....(iii)
∴ Putting value of
m1 and
m2 in Eq. (iii)
−3(‌)=−1 −3b−6=−a+3⇒a−3b−9=0... (iv)
Now, slope of line
CD= slope of
CP m3=‌ ∴ line
Now, solving Eqs. (iv) and (v),
a−3b−9=0⇒3a+b−13=0 Eq. (iv)
×(3) and subtracting Eq. (iv) and Eq. (v),
⇒  
3a−9b−27=0  
‌‌−3a+b−13=0    
‌‌‌‌− 
+  ────────────
   
‌-10b−14=0 ⇒‌‌b=‌⇒b=‌ These value is putting in Eq. (iv)
a+(‌)=0⇒a=‌ Now,
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