Let the three sides of a triangle be on the l | vedclass

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(B) For the first triangle,the lines are $L_1: 4x - 7y + 10 = 0$,$L_2: x + y = 5$,and $L_3: 7x + 4y = 15$. Note that the slopes of $L_1$ and $L_3$ are

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$\sqrt{5}$

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(B) For the first triangle,the lines are $L_1: 4x - 7y + 10 = 0$,$L_2: x + y = 5$,and $L_3: 7x + 4y = 15$. Note that the slopes of $L_1$ and $L_3$ are $m_1 = 4/7$ and $m_3 = -7/4$. Since $m_1 	imes m_3 = -1$,the triangle is a right-angled triangle with the right angle at the intersection of $L_1$ and $L_3$. The orthocentre of a right-angled triangle is the vertex where the right angle is formed. Solving $4x - 7y = -10$ and $7x + 4y = 15$: Multiplying the first by $4$ and second by $7$: $16x - 28y = -40$ and $49x + 28y = 105$. Adding them: $65x = 65 \implies x = 1$. Substituting $x = 1$ in $4(1) - 7y = -10 \implies -7y = -14 \implies y = 2$. So,the orthocentre $H_1$ is $(1, 2)$. For the second triangle formed by $x = 0$,$y = 0$,and $x + y = 1$,it is a right-angled triangle with the right angle at the origin $(0, 0)$. So,the orthocentre $H_2$ is $(0, 0)$. The distance between $H_1(1, 2)$ and $H_2(0, 0)$ is $\sqrt{(1-0)^2 + (2-0)^2} = \sqrt{1 + 4} = \sqrt{5}$.

Let the three sides of a triangle be on the lines $4x - 7y + 10 = 0$,$x + y = 5$,and $7x + 4y = 15$. Then the distance of its orthocentre from the orthocentre of the triangle formed by the lines $x = 0$,$y = 0$,and $x + y = 1$ is

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