If the vertices of a triangles are A(7,−1),B(−2,8) and C(1,2) then which of following is/are correct?
A
Altitude through A is x−2y−9=0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
Altitude through B is 2x−y+12=0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
Altitude through C is x−y+1=0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
Orthocentre is (2,4)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C Altitude through C is x−y+1=0
Let AD,BE and CF be three altitudes of triangle ABC.
Let m1,m2 and m3 be the slopes of AD,BE and CF respectively.
Then, AD⊥BC ⇒ Slope of AD× Slope of BC=−1 ⇒m1×(2−81+2)=−1 ⇒m1=12 BE⊥AC ⇒ Slope of BE× Slope of AC=−1 ⇒m2×(2+11−7)=−1 ⇒m2=2
And. CF⊥AB ⇒ Slope of CF× Slope of AB=−1 ⇒m3×(8+1−2−7)=−1 ⇒m3=1
Since AD passes through A(7,−1) and has slope of m1=12
So, its equation is : y+1=12(x−7) ⇒x−2y−9=0
Similarly, equation of BE is : y−8=2(x+2) ⇒2x−y+12=0
Equation of CF is : y−2=1(x−1) ⇒x−y+1=0
Orthocentre is the intersection of the altitudes. x−2y−9=0...(i) 2x−y+12=0...(ii)
Solving (i) and (ii) we get x=−11,y=−10