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Chapter 14: Problem 1276

Orthocenter of triangle with vertices \((0,0),(3,4)\) and \((4,0)\) is (a) \([3,(5 / 4)]\) (b) \((3,12)\) (c) \([3,(3 / 4)]\) (d) \((3,9)\)

Short Answer

Expert verified
The orthocenter of the given triangle is point (c) \([3,(3 / 4)]\).

Step by step solution

01

Determine the slopes of the sides of the triangle

First, let's find the slopes of each side of the triangle. The slope of line segment \((0, 0)\) to \((3, 4):\) \(m_{1} = \frac{4-0}{3-0} = \frac{4}{3}\) The slope of line segment \((3, 4)\) to \((4, 0):\) \(m_{2} = \frac{0-4}{4-3} = -4\) The slope of line segment \((4, 0)\) to \((0, 0):\) \(m_{3} = \frac{0-0}{4-0} = 0\)
02

Determine the slope of the altitude lines

Now let's find the slopes of the altitude lines. For the altitude line passing through vertex \((0, 0),\) which is perpendicular to the side segment \((3, 4)\) to \((4, 0):\) \(m_{a1} = \frac{-1}{m_{2}} = \frac{1}{4}\) For the altitude line passing through vertex \((3, 4),\) which is perpendicular to the side segment \((4, 0)\) to \((0, 0):\) \(m_{a2} = \frac{-1}{m_{3}} \Rightarrow \text{Since dividing by 0 is undefined, the line is vertical}\)
03

Determine the equation of the altitude lines

Now that we have the slopes of the altitude lines, we can use the point-slope form to find the equations of the lines. The equation of altitude line passing through vertex \((0, 0)\) is: \(y = m_{a1}(x-0)\) \(y = \frac{1}{4}x\) The equation of altitude line passing through vertex \((3, 4)\) is a vertical line: \(x = 3\)
04

Find the point of intersection of the altitude lines

To find the orthocenter, we need to find the point of intersection between the two altitude lines: For \(x = 3\), substitute in the equation of the altitude line passing through vertex \((0, 0),\) \(y = \frac{1}{4}(3)\) \(y = \frac{3}{4}\) So, the orthocenter of the triangle is \((3, \frac{3}{4})\) Therefore, the correct choice is: (c) \([3,(3 / 4)]\)

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Most popular questions from this chapter

If the line \((a+1) x+\left(a^{2}-a-2\right) y+a=0\) is parallel to \(Y-\) axis, then \(\mathrm{a}=\ldots \ldots\) (a) - 1 (b) 2 (c) 3 (d) 1

The equation of line passing through the point of intersection of the lines \(3 \mathrm{x}-2 \mathrm{y}=0\) and \(5 \mathrm{x}+\mathrm{y}-2=0\) and making the angle of measure \(\tan ^{-1}(-5)\) with the positive direction of \(\mathrm{x}\) -axis is \(\ldots \ldots\) (a) \(3 x-2 y=0\) (b) \(5 x+y-2=0\) (c) \(5 \mathrm{x}+\mathrm{y}=0\) (d) \(3 \mathrm{x}+2 \mathrm{y}+1=0\)

If the lines \(\mathrm{x}+(\mathrm{a}-1) \mathrm{y}+1=0\) and \(2 \mathrm{x}+\mathrm{a}^{2} \mathrm{y}-1=0\) are perpendicular then \(\ldots \ldots\) (a) \(|\mathrm{a}|=2\) (b) \(0<\mathrm{a}<1\) (c) \(-1<\mathrm{a}<1\) (d) \(a=-1\)

Find the equation of line passing through the point \((\sqrt{3},-1)\) and at a distance \(\sqrt{2}\) units from the origin. (a) \((\sqrt{3}+1) \mathrm{x}+(\sqrt{3}-1) \mathrm{y}=4\) or \((\sqrt{3}-1) \mathrm{x}-(\sqrt{3}+1) \mathrm{y}=4\) (b) \((\sqrt{3}+1) \mathrm{x}+(\sqrt{3}+1) \mathrm{y}=4\) or \((\sqrt{3}-1) \mathrm{x}+(\sqrt{3}+1) \mathrm{y}=4\) (c) \((\sqrt{3}+1) \mathrm{x}+(\sqrt{3}-1) \mathrm{y}=4\) or \((\sqrt{3}-1) \mathrm{x}+(\sqrt{3}+1) \mathrm{y}=4\) (d) \((\sqrt{3}-1) \mathrm{x}+(\sqrt{3}-1) \mathrm{y}=4\) or \((\sqrt{3}+1) \mathrm{x}+(\sqrt{3}+1) \mathrm{y}=4\)

If \(\mathrm{P}(-1,0), \mathrm{Q}(0,0)\) and \(\mathrm{R}(3,3 \sqrt{3})\) are given points, then the equation of the bisector of \(\angle P Q R\) is \(\ldots \ldots\) (a) \((\sqrt{3} / 2) \mathrm{x}+\mathrm{y}=0\) (b) \(x+(\sqrt{3} / 2) \mathrm{y}=0\) (c) \(\sqrt{3 x}+\mathrm{y}=0\) (d) \(x+\sqrt{3} y=0\)
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