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The following is the schematic diagram of the triangle.

The given lines are $ 4x - 7y + 10 = 0 $ , $ x + y - 5 = 0 $ , $ 7x + 4y - 15 = 0 $ which are sides of the triangle.

We can name the equations as

$ 4x - 7y + 10 = 0 $ ……(i)

$ x + y - 5 = 0 $ ……(ii)

$ 7x + 4y - 15 = 0 $ …….(iii)

We will find the slope of line (i),

$

4x - 7y + 10 = 0\\

7y = 4x + 10\\

y = \dfrac{4}{7}x + \dfrac{{10}}{7}

$

The above equation is in the form of $ y = mx + c $ which is the equation of straight-line having slope $ m $ . Hence this equation has slope $ \dfrac{4}{7} $ .

We will find the slope of line (iii)

$

7x + 4y - 15 = 0\\

4y = - 7x + 15\\

y = - \dfrac{7}{4}x + \dfrac{{15}}{4}

$

The slope of line (iii) is $ - \dfrac{7}{4} $ .

Since the product of slope of line (i) and (iii) is $ - 1 $ . Hence line (i) and (iii) are perpendicular lines. Now to find the orthocentre we will find the point of intersection of these lines. Which will give us an orthocentre.

From equation (i) we can write $ y $ as

$

4x - 7y + 10 = 0\\

4x + 10 = 7y\\

y = \dfrac{{4x + 10}}{7}

$ ……(iv)

Now we will substitute the value of $ y $ in equation (iii) we will get

$

7x + 4\left( {\dfrac{{4x + 10}}{7}} \right) - 15 = 0\\

7x + \dfrac{{16x + 40}}{7} - 15 = 0\\

49x + 16x + 40 - 105 = 0\\

65x - 65 = 0\\

65x = 65\\

x = 1

$

We will substitute 1 for $ x $ in equation (iv) to find $ y $ .

$

y = \dfrac{{4\left( 1 \right) + 10}}{7}\\

y = \dfrac{{14}}{7}\\

y = 2

$

Hence we can say that $ \left( {1,2} \right) $ will be the point of intersection for equation (i) and (iii). Hence we can say that $ \left( {1,2} \right) $ is the orthocentre of the triangle.

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