Question

Orthocentre of the triangle whose sides are given by 4x - 7y + 10 = 0, x + y - 5 = 0 & 7x + 4y - 15 = 0 is

• A. (-1, -2)
• B. (1,- 2)
• C. (-1, 2)
• D. (1, 2)

Answer 1

The correct option is D (1, 2)


Let ABC be the triangle and point O be the orthocentre of the triangle ABC.
Let equation of side AB be 4x - 7y + 10 = 0
$\Rightarrow$ 4x - 7y =-10 . . . . (1)
side BC is x + y = 5 . . . (2)
side AC is 7x + 4y = 15 . . . (3)
Solving equation (1) and (2), we get
$x=\frac{25}{11},y=\frac{30}{11}$
So, point B will be $(\frac{25}{11},\frac{30}{11})$
And,
Solving equation (2) and (3), we get
$x=-\frac{5}{3},y=\frac{20}{3}$
So, point C will be $(-\frac{5}{3},\frac{20}{3})$
And, Solving equation (1) and (3), we get
x = 1, y = 2
So, point A will be (1, 2)
Now, altitude AD is perpendicular to BC, therefore equation of AD is x - y + k = 0
AD is passing through the point A(1, 2) so 1 - 2 + k = 0 $\Rightarrow$ k = 1
Thus, equation of AD = x - y + 1 = 0 . . . (4)
And, altitude BE is perpendicular to AC, therefore equation of BE is 4x - 7y + k = 0
BE is passing through the point $B(\frac{25}{11},\frac{30}{11})so4\times \frac{25}{11}-7\times \frac{30}{11}+k=0\Rightarrow k=10$
Thus, equation of BE = 4x - 7y + 10 = 0 . . . (5)
Solving equation (4) and (5), we get
x = 1 , y = 2
Thus, the coordinate of the orthocentre is (1,2)
Hence option (d) is correct