Question

The co-ordinates of the orthocentre of the triangle bounded by the lines, $4x-7y+10=0;x+y=5$ and $7x+4y=15$ is-

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

Answer 1

Answer: (C)

$(1,2)$

We have,

Equations of lines are

$4x-7y+10=0......\left(1\right)$

$x+y=5......\left(2\right)$

$7x+4y=15......\left(3\right)$

From equation (1) and (3) to and we get,

$Slopeofline\left(1\right)$

$4x-7y+10=0$

$\Rightarrow 7y=4x+10$

$\Rightarrow y=\frac{4}{7}x+\frac{10}{7}$

On comparing that,

$y=m_{1}x+C$

Now, $m_1=\frac{4}{7}$

Slope of line (3)

$7x+4y=15$

$\Rightarrow 4y=-7x+15$

$\Rightarrow y=-\frac{7}{4}x+\frac{15}{4}$

On comparing that,

$y=m_{2}x+C$

So,

$m_2=\frac{-7}{4}$

So,

$m_1\times m_2=-1$

Then both lines are perpendicular.

So,

Point of intersection $=$ Orthocenter of triangle

Now,

Point of intersection lines (1) and (2) are

$4x-7y+10=0$

$7x+4y-15=0$

On solving that,

$(x,y)=(1,2)$

Hence, this is the answer.