Question 216 of 303MediumMCQ
If two vertices of a triangle are (5,−1) and (−2,3) and its orthocentre is at (0,0), then the third vertex is: Solutions · 1
Official solution
Let third vertex (a,b); A(5,−1),B(−2,3), orthocentre H(0,0).
AH⊥BC: slope of AH=−51; slope of BC=a+2b−3:
\[-\frac15\times\frac{b-3}{a+2}=-1 \implies b-3=5(a+2) \quad\text{(i)}\]
BH⊥AC: slope of BH=−23; slope of AC=a−5b+1:
\[-\frac32\times\frac{b+1}{a-5}=-1 \implies 3(b+1)=2(a-5) \implies 3b+3=2a-10 \quad\text{(ii)}\]
From (i): b=5a+13. Substitute into (ii): 3(5a+13)+3=2a−10⟹15a+42=2a−10⟹13a=−52⟹a=−4
b=5(−4)+13=−7
\[\text{Third vertex} = (-4,-7)\] Answer the question to unlock