Question

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find the orthocenter of the triangle with the given vertices.

1. n(-1, 0), p(1, 8), q(5, 0) 13. r(-1, 4), s(5, -2), t(-1, -6)

Explanation:

Problem 12:

Step 1: Identify the type of triangle

First, let's analyze the coordinates of \( N(-1,0) \), \( P(1,8) \), and \( Q(5,0) \). The y - coordinates of \( N \) and \( Q \) are both 0, so the line \( NQ \) is horizontal (parallel to the x - axis). The slope of \( NQ \) is \( m_{NQ}=\frac{0 - 0}{5-(-1)} = 0 \). A line perpendicular to a horizontal line is vertical (parallel to the y - axis). So, the altitude from \( P \) to \( NQ \) is a vertical line. Since \( NQ \) is horizontal, the altitude from \( P \) to \( NQ \) passes through \( P \) and has the same x - coordinate as \( P \) (because it's vertical). The x - coordinate of \( P \) is 1, so the equation of the altitude from \( P \) to \( NQ \) is \( x = 1 \).

Step 2: Find the slope of another side

Now, let's find the slope of \( PQ \). The coordinates of \( P(1,8) \) and \( Q(5,0) \). The slope formula is \( m=\frac{y_2 - y_1}{x_2 - x_1} \). So, \( m_{PQ}=\frac{0 - 8}{5 - 1}=\frac{-8}{4}=-2 \). The slope of the altitude from \( N \) to \( PQ \) is the negative reciprocal of \( m_{PQ} \). The negative reciprocal of - 2 is \( \frac{1}{2} \) (since if \( m_1\times m_2=-1 \), then \( m_2 =-\frac{1}{m_1} \), here \( m_1=-2 \), so \( m_2=\frac{1}{2} \)).

Step 3: Equation of the altitude from \( N \)

Using the point - slope form \( y - y_1=m(x - x_1) \), where \( (x_1,y_1)=N(-1,0) \) and \( m = \frac{1}{2} \). So, \( y-0=\frac{1}{2}(x + 1) \), which simplifies to \( y=\frac{1}{2}x+\frac{1}{2} \).

Step 4: Find the intersection of the two altitudes

We know that one altitude has the equation \( x = 1 \). Substitute \( x = 1 \) into the equation of the altitude from \( N \): \( y=\frac{1}{2}(1)+\frac{1}{2}=\frac{1 + 1}{2}=1 \).

Step 1: Identify the type of triangle

The coordinates of \( R(-1,4) \), \( S(5,-2) \), and \( T(-1,-6) \). The x - coordinates of \( R \) and \( T \) are both - 1, so the line \( RT \) is vertical (parallel to the y - axis). The slope of a vertical line is undefined. A line perpendicular to a vertical line is horizontal (parallel to the x - axis). So, the altitude from \( S \) to \( RT \) is a horizontal line. Since \( RT \) is vertical, the altitude from \( S \) to \( RT \) passes through \( S \) and has the same y - coordinate as \( S \). The y - coordinate of \( S \) is - 2, so the equation of the altitude from \( S \) to \( RT \) is \( y=-2 \).

Step 2: Find the slope of another side

Let's find the slope of \( RS \). The coordinates of \( R(-1,4) \) and \( S(5,-2) \). Using the slope formula \( m=\frac{y_2 - y_1}{x_2 - x_1} \), we have \( m_{RS}=\frac{-2 - 4}{5-(-1)}=\frac{-6}{6}=-1 \). The slope of the altitude from \( T \) to \( RS \) is the negative reciprocal of \( m_{RS} \). The negative reciprocal of - 1 is 1 (since \( (-1)\times1=-1 \)).

Step 3: Equation of the altitude from \( T \)

Using the point - slope form \( y - y_1=m(x - x_1) \), where \( (x_1,y_1)=T(-1,-6) \) and \( m = 1 \). So, \( y+6=1\times(x + 1) \), which simplifies to \( y=x + 1-6=x - 5 \).

Step 4: Find the intersection of the two altitudes

We know that one altitude has the equation \( y=-2 \). Substitute \( y = - 2 \) into the equation of the altitude from \( T \): \( -2=x - 5 \). Solving for \( x \), we add 5 to both sides: \( x=-2 + 5=3 \).

Answer:

The orthocenter of triangle \( NPQ \) is \( (1,1) \)

Problem 13: