The coordinates of the orthocentre of the tri | vedclass

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(D) The sides of the triangle are $L_1: x = 3$,$L_2: y = 4$,and $L_3: 3x + 4y = 6$. First,find the vertices of the triangle by solving the intersectio

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$(3, 4)$

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(D) The sides of the triangle are $L_1: x = 3$,$L_2: y = 4$,and $L_3: 3x + 4y = 6$. First,find the vertices of the triangle by solving the intersection of these lines: $1$. Intersection of $L_1$ and $L_2$: $x = 3$ and $y = 4$ gives the vertex $A(3, 4)$. $2$. Intersection of $L_1$ and $L_3$: $x = 3$ and $3(3) + 4y = 6 \implies 9 + 4y = 6 \implies 4y = -3 \implies y = -3/4$. So,vertex $B(3, -3/4)$. $3$. Intersection of $L_2$ and $L_3$: $y = 4$ and $3x + 4(4) = 6 \implies 3x + 16 = 6 \implies 3x = -10 \implies x = -10/3$. So,vertex $C(-10/3, 4)$. Since $L_1$ is a vertical line $(x = 3)$ and $L_2$ is a horizontal line $(y = 4)$,they are perpendicular to each other. Therefore,the triangle is a right-angled triangle with the right angle at the intersection of $x = 3$ and $y = 4$,which is $(3, 4)$. In a right-angled triangle,the orthocentre is the vertex where the right angle is located. Thus,the orthocentre is $(3, 4)$.

The coordinates of the orthocentre of the triangle whose sides are given by the lines $x = 3$,$y = 4$,and $3x + 4y = 6$ are:

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