Question

If the equations of the sides of a triangle are 7x + y - 10 =0, x-2y + 5=0 and x+ y + 2 = 0, find the orthocenter of the triangle.
triangle ​

Answer 1

Answer:

If the equation of the sides of a triangle are 7x+y−10=0;x−2y+5=0;x+y+2=0 find orthocentre of the triangle.

Since z seems to be a typing error I take its nearest looking 2

I shall not find the vertices.

L1::7x+y−10=0(1)

L2::x−2y+5=0(2)

L3::x+y+2=0(3)

(A)

A line passing through intersection of L1 and L2 is given by

7x+y−10+λ(x−2y+5)=0

(7+λ)x+(1−2λ)y−10+5λ=0(4)

If this is to pass though ortocenter it must be perpendicular to L_3

i.e. 1* (7+λ)+1∗(1−2λ)=0

8- λ)=0orλ=8

Back substituting in (4)

x−y+2=0(5)

(B)

A line passing through intersection of L1 and L3 is given by

7x+y−10+μ(x+y+2)=0

(7+μ)x+(1+μ)y−10+2μ=0(6)

If this is to pass though ortocenter it must be perpendicular to L_2

i.e. 1* (7+μ)−2∗(1+μ)=0

5- μ=0orμ=5

Back substituting in (6)

2x+y=0(7)

Now the orthocenter is the intersection of (5) and (7)

Solving x=−2/3 and y=4/3

So orthocenter is (−2/3,4/3)