The coordinates of the orthocentre of the tri | vedclass

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(C) Let the lines be $L_1: 4x - 7y + 10 = 0$,$L_2: x + y - 5 = 0$,and $L_3: 7x + 4y - 15 = 0$.First,check the slopes of the lines:Slope of $L_1$ $(m_1

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$(1, 2)$

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(C) Let the lines be $L_1: 4x - 7y + 10 = 0$,$L_2: x + y - 5 = 0$,and $L_3: 7x + 4y - 15 = 0$.First,check the slopes of the lines:Slope of $L_1$ $(m_1)$ = $4/7$.Slope of $L_2$ $(m_2)$ = $-1$.Slope of $L_3$ $(m_3)$ = $-7/4$.Since $m_1 	imes m_3 = (4/7) 	imes (-7/4) = -1$,the lines $L_1$ and $L_3$ are perpendicular to each other.Therefore,the triangle formed by these lines 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 $L_1$ and $L_3$:$4x - 7y = -10$ $(i)$$7x + 4y = 15$ (ii)Multiply $(i)$ by $4$ and (ii) by $7$:$16x - 28y = -40$$49x + 28y = 105$Adding these equations: $65x = 65 \Rightarrow x = 1$.Substituting $x = 1$ into $(i)$: $4(1) - 7y = -10$ $\Rightarrow -7y = -14$ $\Rightarrow y = 2$.The intersection point is $(1, 2)$,which is the orthocentre.

The coordinates of the orthocentre of the triangle bounded by the lines $4x - 7y + 10 = 0$,$x + y = 5$,and $7x + 4y = 15$ are:

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