Log out Go to App
Go to App

Learning Materials

Features

Discover

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) \([3,(3 / 4)]\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A line intersects \(\mathrm{X}\) -axis and Y-axis at \(\mathrm{A}\) and \(\mathrm{B}\) respectively. If \(\mathrm{AB}=15\) and \(\underline{\mathrm{AB}}\) makes a triangle of area 54 units with coordinate axes, then the equation of \(\underline{A B}\) is \(\ldots .\) (a) \(4 x \pm 3 y=36\) or \(3 x \pm 4 y=36\) (b) \(4 x \pm 3 y=24\) or \(3 x \pm 4 y=24\) (c) \(-4 \mathrm{x} \pm 3 \mathrm{y}=24\) or \(-3 \mathrm{x} \pm 4 \mathrm{y}=24\) (d) \(-4 x \pm 3 y=12\) or \(-3 x \pm 4 y=12\)

Four points \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right),\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right),\left(\mathrm{x}_{3}, \mathrm{y}_{3}\right)\) and \(\left(\mathrm{x}_{4}, \mathrm{y}_{4}\right)\) are such that \({ }^{4} \sum_{\mathrm{i}=1}\left(\mathrm{x}_{\mathrm{i}}^{2}+\mathrm{y}_{\mathrm{i}}^{2}\right) \leq 2\left(\mathrm{x}_{1} \mathrm{x}_{3}+\mathrm{x}_{2} \mathrm{x}_{4}+\mathrm{y}_{1} \mathrm{y}_{2}+\mathrm{y}_{3} \mathrm{y}_{4}\right)\). Then these points are vertices of (a) Parallelogram (b) Rectangle (c) Square (d) Rhombus

If a vertex of a triangle is \((1,1)\) and the mid-points of two sides through this vertex are \((-1,2)\) and \((3,2)\), then centroid of the triangle is (a) \([(1 / 3),(7 / 3)]\) (b) \([1,(7 / 3)]\) (c) \([-(1 / 3),(7 / 3)]\) (d) \([-1,(7 / 3)]\)

The equation of the lines with slope \(-2\) and intersecting \(\mathrm{x}\) -axis at points distance 3 unit from the origin is...... (a) \(2 \mathrm{x}+\mathrm{y} \pm 6=0\) (b) \(x+2 y \pm 6=0\) (c) \(2 \mathrm{x}+\mathrm{y} \pm 3=0\) (d) \(x+2 y \pm 3=0\)

The equation of a line at a distance \(\sqrt{5}\) units from the origin and the ratio of the intercepts on the axes is \(1: 2\), is \(\ldots \ldots\) (a) \(2 \mathrm{x}+\mathrm{y} \pm 5=0\) (b) \(2 x+y \pm 5=0\) (c) \(x-2 y \pm 5=0\) (d) None of these
See all solutions

Recommended explanations on English Textbooks

Rhetoric

Read Explanation

Semiotics

Read Explanation

Text Comparison

Read Explanation

Morphology

Read Explanation

Email

Read Explanation

Blog

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free