Question

Orthocentre of a triangle whose sides are given by 4x-7y+10= 0 ,x+y-5=0 and 7x+4y-a5=0 is ??a) (-1, -2) b) (1, -2) c) (-1, 2)d) (1, 2)

Answer 1

Answer:

option (d) is  correct

Step-by-step explanation:

Sides of the given triangle are

4x-7y+10= 0, x+y-5=0 and 7x+4y-15=0

$\text{Slope of 4x-7y+10= 0 is }m_1=\frac{4}{7}$

$\text{Slope of 7x+4y-15= 0 is }m_2=\frac{-7}{4}$

$m_1{\times}m_2=\frac{4}{7}{\times}\frac{-7}{4}=-1$

$\therefore\text{The lines are perpendicular}$

$\text{Hence, the given triangle is right angled triangle}$

We know that in a right angled triangle the point of intersection of the perpendicular sides is its orthocentre

To find orthocentre:

4x-7y+10= 0 and 7x+4y-15=0

16x-28y+40=0

49x+28y-105=0

65x=65

x=1

put x=1 in 7x+4y-15=0

7+4y-15=0

4y=8

y=2

$\therefore\textbf{Orthocentre is (1,2)}$

Answer 2

Answer : (1,2).

Explanation :

4x+7y+10=0

x+y-5=0

7x+4y-15=0