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Question: The orthocentre of the triangle formed by \[\left( {0,0} \right),\left( {8,0} \right),\left( {4,6} \...

The orthocentre of the triangle formed by (0,0),(8,0),(4,6)\left( {0,0} \right),\left( {8,0} \right),\left( {4,6} \right) is
A) (4,83)\left( {4,\dfrac{8}{3}} \right)
B) (3,4)(3,4)
C) (4,3)(4,3)
D) (3,4)( - 3,4)

Explanation

Solution

For solving this particular question, we must know that slope of a line can also be found if two points on the line are given . let the two points on the line be (x1,y1),(x2,y2)({x_1},{y_1}),({x_2},{y_2}) respectively.
Then the slope is given by , m=y2y1x2x1m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} .
Slope is also defined as the ratio of change in yy over the change in xxbetween any two points.

Complete step-by-step solution:
It is given that (0,0),(8,0),(4,6)\left( {0,0} \right),\left( {8,0} \right),\left( {4,6} \right) are the vertices of a triangle.
Let ABC be a triangle having the vertices (0,0),(8,0),(4,6)\left( {0,0} \right),\left( {8,0} \right),\left( {4,6} \right) ,
Let ‘A’ be a vertex (0,0)\left( {0,0} \right) ,
‘B’ be a vertex (8,0)\left( {8,0} \right) ,
And ‘C’ be a vertex (4,6)\left( {4,6} \right) .
Slope of a line can also be found if two points on the line are given . let the two points on the line be (x1,y1),(x2,y2)({x_1},{y_1}),({x_2},{y_2}) respectively.
Then slope is given by , m=y2y1x2x1m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} .
Slope is also defined as the ratio of change in yy over the change in xxbetween any two points.
Therefore, slope of BC=6048=23BC = \dfrac{{6 - 0}}{{4 - 8}} = - \dfrac{2}{3} .
Now, we have to find equation of the line through ‘A’ which is perpendicular to BCBC is ,
y0=(23)(x0) 2x3y=0......................(1)  \Rightarrow y - 0 = \left( {\dfrac{2}{3}} \right)(x - 0) \\\ \Rightarrow 2x - 3y = 0......................(1) \\\
Now, slope of CA=0604=32CA = \dfrac{{0 - 6}}{{0 - 4}} = \dfrac{3}{2} .
Now, we have to find equation of the line through ‘B’ which is perpendicular to CACA is ,
y0=(32)(x8) 2x+3y=16.....................(2)  \Rightarrow y - 0 = \left( { - \dfrac{3}{2}} \right)(x - 8) \\\ \Rightarrow 2x + 3y = 16.....................(2) \\\
Now, we have to solve (1),(2)(1),(2) , after solving we will get the orthocentre as (4,83)\left( {4,\dfrac{8}{3}} \right) .

Therefore, option ‘A’ is the correct option.

Note: This type of linear equations sometimes called slope-intercept form because we can easily find the slope and the intercept of the corresponding lines. This also allows us to graph it. We can quickly tell the slope i.e., mm the y-intercepts i.e., (y,0)(y,0) and the x-intercept i.e., (0,y)(0,y) .we can graph the corresponding line .